CBSE Class 10 Probability Sure Shot Questions Set 02

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PRACTICE QUESTIONS
CLASS X : CHAPTER - 15
PROBABILITY
PLAYING CARDS BASED QUESTIONS

1. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(a) an ace card

Answer:
1. Favourable cards: 4 out of 52 cards left in the deck.
2. P = \( \frac{4}{52} \) = \( \frac{1}{13} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(b) a red card

Answer:
1. Favourable cards: 26 out of 52 cards left in the deck.
2. P = \( \frac{26}{52} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(c) either red or king card

Answer:
1. Favourable cards: 28 out of 52 cards left in the deck.
2. P = \( \frac{28}{52} \) = \( \frac{7}{13} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 28 favourable cards in total out of 52 cards left.

(d) red and a king

Answer:
1. Favourable cards: 2 out of 52 cards left in the deck.
2. P = \( \frac{2}{52} \) = \( \frac{1}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(e) '2' of spades

Answer:
1. Favourable cards: 1 out of 52 cards left in the deck.
2. P = \( \frac{1}{52} \)

Teacher's Note:
a) Only one card in the deck matches, so the numerator is 1.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(f) '10' of a black suit

Answer:
1. Favourable cards: 2 out of 52 cards left in the deck.
2. P = \( \frac{2}{52} \) = \( \frac{1}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(g) a queen of black suit

Answer:
1. Favourable cards: 2 out of 52 cards left in the deck.
2. P = \( \frac{2}{52} \) = \( \frac{1}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(h) either a black card or a king

Answer:
1. Favourable cards: 28 out of 52 cards left in the deck.
2. P = \( \frac{28}{52} \) = \( \frac{7}{13} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 28 favourable cards in total out of 52 cards left.

(i) black and king card

Answer:
1. Favourable cards: 2 out of 52 cards left in the deck.
2. P = \( \frac{2}{52} \) = \( \frac{1}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(j) a jack, queen or a king

Answer:
1. Favourable cards: 12 out of 52 cards left in the deck.
2. P = \( \frac{12}{52} \) = \( \frac{3}{13} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(k) a heart card

Answer:
1. Favourable cards: 13 out of 52 cards left in the deck.
2. P = \( \frac{13}{52} \) = \( \frac{1}{4} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(l) a queen card

Answer:
1. Favourable cards: 4 out of 52 cards left in the deck.
2. P = \( \frac{4}{52} \) = \( \frac{1}{13} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(m) the ace of spades

Answer:
1. Favourable cards: 1 out of 52 cards left in the deck.
2. P = \( \frac{1}{52} \)

Teacher's Note:
a) Only one card in the deck matches, so the numerator is 1.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(n) the seven of clubs

Answer:
1. Favourable cards: 1 out of 52 cards left in the deck.
2. P = \( \frac{1}{52} \)

Teacher's Note:
a) Only one card in the deck matches, so the numerator is 1.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(o) a ten

Answer:
1. Favourable cards: 4 out of 52 cards left in the deck.
2. P = \( \frac{4}{52} \) = \( \frac{1}{13} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(p) a black card

Answer:
1. Favourable cards: 26 out of 52 cards left in the deck.
2. P = \( \frac{26}{52} \) = \( \frac{1}{2} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(q) neither a heart nor a king

Answer:
1. Favourable cards: 36 out of 52 cards left in the deck.
2. P = \( \frac{36}{52} \) = \( \frac{9}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 36 favourable cards out of 52, so the probability is \( \frac{9}{13} \).

(r) neither an ace nor a king

Answer:
1. Favourable cards: 44 out of 52 cards left in the deck.
2. P = \( \frac{44}{52} \) = \( \frac{11}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 44 favourable cards out of 52, so the probability is \( \frac{11}{13} \).

(s) neither a red card nor a queen card

Answer:
1. Favourable cards: 24 out of 52 cards left in the deck.
2. P = \( \frac{24}{52} \) = \( \frac{6}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 24 favourable cards out of 52, so the probability is \( \frac{6}{13} \).

(t) a face card or an ace

Answer:
1. Favourable cards: 16 out of 52 cards left in the deck.
2. P = \( \frac{16}{52} \) = \( \frac{4}{13} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 16 favourable cards in total out of 52 cards left.

(u) a face card or a black card

Answer:
1. Favourable cards: 32 out of 52 cards left in the deck.
2. P = \( \frac{32}{52} \) = \( \frac{8}{13} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 32 favourable cards in total out of 52 cards left.

(v) a face card and a black card

Answer:
1. Favourable cards: 6 out of 52 cards left in the deck.
2. P = \( \frac{6}{52} \) = \( \frac{3}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(w) neither a face card nor an ace

Answer:
1. Favourable cards: 36 out of 52 cards left in the deck.
2. P = \( \frac{36}{52} \) = \( \frac{9}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 36 favourable cards out of 52, so the probability is \( \frac{9}{13} \).

(x) neither a face card nor a '10' card

Answer:
1. Favourable cards: 36 out of 52 cards left in the deck.
2. P = \( \frac{36}{52} \) = \( \frac{9}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 36 favourable cards out of 52, so the probability is \( \frac{9}{13} \).

(y) either a king or red card

Answer:
1. Favourable cards: 28 out of 52 cards left in the deck.
2. P = \( \frac{28}{52} \) = \( \frac{7}{13} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 28 favourable cards in total out of 52 cards left.

(z) either an ace or black card

Answer:
1. Favourable cards: 28 out of 52 cards left in the deck.
2. P = \( \frac{28}{52} \) = \( \frac{7}{13} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 28 favourable cards in total out of 52 cards left.

(aa) an ace and a black card

Answer:
1. Favourable cards: 2 out of 52 cards left in the deck.
2. P = \( \frac{2}{52} \) = \( \frac{1}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(bb) a king of red colour card

Answer:
1. Favourable cards: 2 out of 52 cards left in the deck.
2. P = \( \frac{2}{52} \) = \( \frac{1}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(cc) a face card

Answer:
1. Favourable cards: 12 out of 52 cards left in the deck.
2. P = \( \frac{12}{52} \) = \( \frac{3}{13} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(dd) a red face card

Answer:
1. Favourable cards: 6 out of 52 cards left in the deck.
2. P = \( \frac{6}{52} \) = \( \frac{3}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(ee) the jack of hearts

Answer:
1. Favourable cards: 1 out of 52 cards left in the deck.
2. P = \( \frac{1}{52} \)

Teacher's Note:
a) Only one card in the deck matches, so the numerator is 1.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(ff) a spade card

Answer:
1. Favourable cards: 13 out of 52 cards left in the deck.
2. P = \( \frac{13}{52} \) = \( \frac{1}{4} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(gg) the queen of diamonds

Answer:
1. Favourable cards: 1 out of 52 cards left in the deck.
2. P = \( \frac{1}{52} \)

Teacher's Note:
a) Only one card in the deck matches, so the numerator is 1.
b) The full deck has 52 cards, so the denominator is 52 for this question.

(hh) '9' of black suit

Answer:
1. Favourable cards: 2 out of 52 cards left in the deck.
2. P = \( \frac{2}{52} \) = \( \frac{1}{26} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(ii) a face card or spade card

Answer:
1. Favourable cards: 22 out of 52 cards left in the deck.
2. P = \( \frac{22}{52} \) = \( \frac{11}{26} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 22 favourable cards in total out of 52 cards left.

2. Five cards - the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is
(a) an ace?
(b) a queen?

(i) What is the probability that the card is the queen?

Answer:
1. Favourable cards: 1 (the queen of diamonds) out of 5 cards.
2. P = \( \frac{1}{5} \)

Teacher's Note:
a) There is exactly one queen among the five cards, so the numerator is 1.
b) Each of the five cards is equally likely to be picked.

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is

(a) an ace?

Answer:
1. Cards left after the queen is put aside: 4 (ten, jack, king and ace of diamonds).
2. Favourable cards: 1 (the ace of diamonds).
3. P = \( \frac{1}{4} \)

Teacher's Note:
a) The queen is removed from the deck, so the total changes from 5 to 4.
b) Only the ace of diamonds is an ace among the four remaining cards.

(b) a queen?

Answer: P = 0, since the only queen has already been put aside.

Teacher's Note:
a) The queen was removed in part (ii), so no queen remains in the four cards.
b) An impossible event always has probability 0.

3. The king , queen and jack of clubs are removed from a pack of 52 playing cards. One card is selected at random from the remaining cards. Find the probability that the card is

(a) neither a heart nor a king

Answer:
1. Favourable cards: 34 out of 49 cards left in the deck.
2. P = \( \frac{34}{49} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 34 favourable cards out of 49, so the probability is \( \frac{34}{49} \).

(b) neither an ace nor a king

Answer:
1. Favourable cards: 42 out of 49 cards left in the deck.
2. P = \( \frac{42}{49} \) = \( \frac{6}{7} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 42 favourable cards out of 49, so the probability is \( \frac{6}{7} \).

(c) neither a red card nor a queen card

Answer:
1. Favourable cards: 22 out of 49 cards left in the deck.
2. P = \( \frac{22}{49} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 22 favourable cards out of 49, so the probability is \( \frac{22}{49} \).

(d) a black card or an ace

Answer:
1. Favourable cards: 25 out of 49 cards left in the deck.
2. P = \( \frac{25}{49} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 25 favourable cards in total out of 49 cards left.

(e) either a heart or a spade card

Answer:
1. Favourable cards: 26 out of 49 cards left in the deck.
2. P = \( \frac{26}{49} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 26 favourable cards in total out of 49 cards left.

(f) a king card

Answer:
1. Favourable cards: 3 out of 49 cards left in the deck.
2. P = \( \frac{3}{49} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(g) a heart card

Answer:
1. Favourable cards: 13 out of 49 cards left in the deck.
2. P = \( \frac{13}{49} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(h) a red card

Answer:
1. Favourable cards: 26 out of 49 cards left in the deck.
2. P = \( \frac{26}{49} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(i) a black card

Answer:
1. Favourable cards: 23 out of 49 cards left in the deck.
2. P = \( \frac{23}{49} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(j) a spade card

Answer:
1. Favourable cards: 13 out of 49 cards left in the deck.
2. P = \( \frac{13}{49} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(k) a diamond card

Answer:
1. Favourable cards: 13 out of 49 cards left in the deck.
2. P = \( \frac{13}{49} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(l) a club card

Answer:
1. Favourable cards: 10 out of 49 cards left in the deck.
2. P = \( \frac{10}{49} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(m) either an ace card or black card

Answer:
1. Favourable cards: 25 out of 49 cards left in the deck.
2. P = \( \frac{25}{49} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 25 favourable cards in total out of 49 cards left.

(n) an ace card

Answer:
1. Favourable cards: 4 out of 49 cards left in the deck.
2. P = \( \frac{4}{49} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(o) a face card

Answer:
1. Favourable cards: 9 out of 49 cards left in the deck.
2. P = \( \frac{9}{49} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 49 cards after removal, so the denominator is 49, not 52.

(p) a face card with red colour

Answer:
1. Favourable cards: 6 out of 49 cards left in the deck.
2. P = \( \frac{6}{49} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(q) neither '10' card nor an ace

Answer:
1. Favourable cards: 41 out of 49 cards left in the deck.
2. P = \( \frac{41}{49} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 41 favourable cards out of 49, so the probability is \( \frac{41}{49} \).

(r) an even number card

Answer:
1. Favourable cards: 20 out of 49 cards left in the deck.
2. P = \( \frac{20}{49} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 49 cards left, which gives 20.

(s) an odd number card

Answer:
1. Favourable cards: 16 out of 49 cards left in the deck.
2. P = \( \frac{16}{49} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 49 cards left, which gives 16.

(t) not a natural number

Answer:
1. Favourable cards: 9 out of 49 cards left in the deck.
2. P = \( \frac{9}{49} \)

Teacher's Note:
a) An ace counts as 1 and number cards from 2 to 10 are natural numbers, but jack, queen and king are not natural numbers.
b) So the favourable cards are the face cards left in the deck, which number 9 out of 49.

6. All spades are removed from a well shuffled deck of 52 cards and then one card is drawn randomly from the remaining cards. Find the probability of getting

(a) neither a heart nor a king

Answer:
1. Favourable cards: 24 out of 39 cards left in the deck.
2. P = \( \frac{24}{39} \) = \( \frac{8}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 24 favourable cards out of 39, so the probability is \( \frac{8}{13} \).

(b) neither an ace nor a king

Answer:
1. Favourable cards: 33 out of 39 cards left in the deck.
2. P = \( \frac{33}{39} \) = \( \frac{11}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 33 favourable cards out of 39, so the probability is \( \frac{11}{13} \).

(c) neither a red card nor a queen card

Answer:
1. Favourable cards: 12 out of 39 cards left in the deck.
2. P = \( \frac{12}{39} \) = \( \frac{4}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 12 favourable cards out of 39, so the probability is \( \frac{4}{13} \).

(d) a black card or an ace

Answer:
1. Favourable cards: 15 out of 39 cards left in the deck.
2. P = \( \frac{15}{39} \) = \( \frac{5}{13} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 15 favourable cards in total out of 39 cards left.

(e) either a heart or a spade card

Answer:
1. Favourable cards: 13 out of 39 cards left in the deck.
2. P = \( \frac{13}{39} \) = \( \frac{1}{3} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 13 favourable cards in total out of 39 cards left.

(f) a red card

Answer:
1. Favourable cards: 26 out of 39 cards left in the deck.
2. P = \( \frac{26}{39} \) = \( \frac{2}{3} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 39 cards after removal, so the denominator is 39, not 52.

(g) a black card

Answer:
1. Favourable cards: 13 out of 39 cards left in the deck.
2. P = \( \frac{13}{39} \) = \( \frac{1}{3} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 39 cards after removal, so the denominator is 39, not 52.

(h) a spade card

Answer:
1. Favourable cards: 0 out of 39 cards left in the deck.
2. P = 0

Teacher's Note:
a) No card in the deck satisfies this condition, so the event is impossible and its probability is 0.
b) Check that the removed cards have not left any card of this type in the deck.

(i) a diamond card

Answer:
1. Favourable cards: 13 out of 39 cards left in the deck.
2. P = \( \frac{13}{39} \) = \( \frac{1}{3} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 39 cards after removal, so the denominator is 39, not 52.

(j) a club card

Answer:
1. Favourable cards: 13 out of 39 cards left in the deck.
2. P = \( \frac{13}{39} \) = \( \frac{1}{3} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 39 cards after removal, so the denominator is 39, not 52.

(k) either an ace card or black card

Answer:
1. Favourable cards: 15 out of 39 cards left in the deck.
2. P = \( \frac{15}{39} \) = \( \frac{5}{13} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 15 favourable cards in total out of 39 cards left.

(l) an ace card

Answer:
1. Favourable cards: 3 out of 39 cards left in the deck.
2. P = \( \frac{3}{39} \) = \( \frac{1}{13} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 39 cards after removal, so the denominator is 39, not 52.

(m) a face card with red colour

Answer:
1. Favourable cards: 6 out of 39 cards left in the deck.
2. P = \( \frac{6}{39} \) = \( \frac{2}{13} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(n) neither '10' card nor an ace

Answer:
1. Favourable cards: 33 out of 39 cards left in the deck.
2. P = \( \frac{33}{39} \) = \( \frac{11}{13} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 33 favourable cards out of 39, so the probability is \( \frac{11}{13} \).

(o) an even number card

Answer:
1. Favourable cards: 15 out of 39 cards left in the deck.
2. P = \( \frac{15}{39} \) = \( \frac{5}{13} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 39 cards left, which gives 15.

(p) a face card

Answer:
1. Favourable cards: 9 out of 39 cards left in the deck.
2. P = \( \frac{9}{39} \) = \( \frac{3}{13} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 39 cards after removal, so the denominator is 39, not 52.

(q) an odd number card

Answer:
1. Favourable cards: 12 out of 39 cards left in the deck.
2. P = \( \frac{12}{39} \) = \( \frac{4}{13} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 39 cards left, which gives 12.

4. All face cards are removed from a well shuffled deck of 52 cards and then one card is drawn randomly from the remaining cards. Find the probability of getting

(a) neither a heart nor a king

Answer:
1. Favourable cards: 30 out of 40 cards left in the deck.
2. P = \( \frac{30}{40} \) = \( \frac{3}{4} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 30 favourable cards out of 40, so the probability is \( \frac{3}{4} \).

(b) neither an ace nor a king

Answer:
1. Favourable cards: 36 out of 40 cards left in the deck.
2. P = \( \frac{36}{40} \) = \( \frac{9}{10} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 36 favourable cards out of 40, so the probability is \( \frac{9}{10} \).

(c) neither a red card nor a queen card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 20 favourable cards out of 40, so the probability is \( \frac{1}{2} \).

(d) a black card or an ace

Answer:
1. Favourable cards: 22 out of 40 cards left in the deck.
2. P = \( \frac{22}{40} \) = \( \frac{11}{20} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 22 favourable cards in total out of 40 cards left.

(e) either a heart or a spade card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 20 favourable cards in total out of 40 cards left.

(f) a red card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(g) a black card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(h) a spade card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(i) a diamond card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(j) a club card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(k) either an ace card or black card

Answer:
1. Favourable cards: 22 out of 40 cards left in the deck.
2. P = \( \frac{22}{40} \) = \( \frac{11}{20} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 22 favourable cards in total out of 40 cards left.

(l) an ace card

Answer:
1. Favourable cards: 4 out of 40 cards left in the deck.
2. P = \( \frac{4}{40} \) = \( \frac{1}{10} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(m) a face card with red colour

Answer:
1. Favourable cards: 0 out of 40 cards left in the deck.
2. P = 0

Teacher's Note:
a) No card in the deck satisfies this condition, so the event is impossible and its probability is 0.
b) Check that the removed cards have not left any card of this type in the deck.

(n) neither '10' card nor an ace

Answer:
1. Favourable cards: 32 out of 40 cards left in the deck.
2. P = \( \frac{32}{40} \) = \( \frac{4}{5} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 32 favourable cards out of 40, so the probability is \( \frac{4}{5} \).

(o) an even number card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 40 cards left, which gives 20.

(p) an odd number card

Answer:
1. Favourable cards: 16 out of 40 cards left in the deck.
2. P = \( \frac{16}{40} \) = \( \frac{2}{5} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 40 cards left, which gives 16.

5. All cards of ace, jack and queen are removed from a deck of playing cards. One card is drawn at random from the remaining cards, find the probability that the card drawn

(a) neither a heart nor a king

Answer:
1. Favourable cards: 27 out of 40 cards left in the deck.
2. P = \( \frac{27}{40} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 27 favourable cards out of 40, so the probability is \( \frac{27}{40} \).

(b) neither an ace nor a king

Answer:
1. Favourable cards: 36 out of 40 cards left in the deck.
2. P = \( \frac{36}{40} \) = \( \frac{9}{10} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 36 favourable cards out of 40, so the probability is \( \frac{9}{10} \).

(c) neither a red card nor a queen card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 20 favourable cards out of 40, so the probability is \( \frac{1}{2} \).

(d) a black card or an ace

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 20 favourable cards in total out of 40 cards left.

(e) either a heart or a spade card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 20 favourable cards in total out of 40 cards left.

(f) a king card

Answer:
1. Favourable cards: 4 out of 40 cards left in the deck.
2. P = \( \frac{4}{40} \) = \( \frac{1}{10} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(g) a heart card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(h) a red card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(i) a black card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(j) a spade card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(k) a diamond card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(l) a club card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(m) either an ace card or black card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 20 favourable cards in total out of 40 cards left.

(n) an ace card

Answer:
1. Favourable cards: 0 out of 40 cards left in the deck.
2. P = 0

Teacher's Note:
a) No card in the deck satisfies this condition, so the event is impossible and its probability is 0.
b) Check that the removed cards have not left any card of this type in the deck.

(o) a face card

Answer:
1. Favourable cards: 4 out of 40 cards left in the deck.
2. P = \( \frac{4}{40} \) = \( \frac{1}{10} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(p) a face card with red colour

Answer:
1. Favourable cards: 2 out of 40 cards left in the deck.
2. P = \( \frac{2}{40} \) = \( \frac{1}{20} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(q) neither '10' card nor an ace

Answer:
1. Favourable cards: 36 out of 40 cards left in the deck.
2. P = \( \frac{36}{40} \) = \( \frac{9}{10} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 36 favourable cards out of 40, so the probability is \( \frac{9}{10} \).

(r) an even number card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 40 cards left, which gives 20.

(s) an odd number card

Answer:
1. Favourable cards: 16 out of 40 cards left in the deck.
2. P = \( \frac{16}{40} \) = \( \frac{2}{5} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 40 cards left, which gives 16.

(t) not a natural number

Answer:
1. Favourable cards: 4 out of 40 cards left in the deck.
2. P = \( \frac{4}{40} \) = \( \frac{1}{10} \)

Teacher's Note:
a) An ace counts as 1 and number cards from 2 to 10 are natural numbers, but jack, queen and king are not natural numbers.
b) So the favourable cards are the face cards left in the deck, which number 4 out of 40.

6. All cards of '10', an ace and queen cards are removed from a well shuffled deck of 52 cards and then one card is drawn randomly from the remaining cards. Find the probability of getting

(a) neither a heart nor a king

Answer:
1. Favourable cards: 27 out of 40 cards left in the deck.
2. P = \( \frac{27}{40} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 27 favourable cards out of 40, so the probability is \( \frac{27}{40} \).

(b) neither an ace nor a king

Answer:
1. Favourable cards: 36 out of 40 cards left in the deck.
2. P = \( \frac{36}{40} \) = \( \frac{9}{10} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 36 favourable cards out of 40, so the probability is \( \frac{9}{10} \).

(c) neither a red card nor a queen card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 20 favourable cards out of 40, so the probability is \( \frac{1}{2} \).

(d) a black card or an ace

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 20 favourable cards in total out of 40 cards left.

(e) either a heart or a spade card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 20 favourable cards in total out of 40 cards left.

(f) a king card

Answer:
1. Favourable cards: 4 out of 40 cards left in the deck.
2. P = \( \frac{4}{40} \) = \( \frac{1}{10} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(g) a heart card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(h) a red card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(i) a black card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(j) a spade card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(k) a diamond card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(l) a club card

Answer:
1. Favourable cards: 10 out of 40 cards left in the deck.
2. P = \( \frac{10}{40} \) = \( \frac{1}{4} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(m) either an ace card or black card

Answer:
1. Favourable cards: 20 out of 40 cards left in the deck.
2. P = \( \frac{20}{40} \) = \( \frac{1}{2} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 20 favourable cards in total out of 40 cards left.

(n) an ace card

Answer:
1. Favourable cards: 0 out of 40 cards left in the deck.
2. P = 0

Teacher's Note:
a) No card in the deck satisfies this condition, so the event is impossible and its probability is 0.
b) Check that the removed cards have not left any card of this type in the deck.

(o) a face card

Answer:
1. Favourable cards: 8 out of 40 cards left in the deck.
2. P = \( \frac{8}{40} \) = \( \frac{1}{5} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 40 cards after removal, so the denominator is 40, not 52.

(p) a face card with red colour

Answer:
1. Favourable cards: 4 out of 40 cards left in the deck.
2. P = \( \frac{4}{40} \) = \( \frac{1}{10} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(q) neither '10' card nor an ace

Answer:
1. Favourable cards: 40 out of 40 cards left in the deck.
2. P = \( \frac{40}{40} \) = 1

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 40 favourable cards out of 40, so the probability is 1.

(r) an even number card

Answer:
1. Favourable cards: 16 out of 40 cards left in the deck.
2. P = \( \frac{16}{40} \) = \( \frac{2}{5} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 40 cards left, which gives 16.

(s) an odd number card

Answer:
1. Favourable cards: 16 out of 40 cards left in the deck.
2. P = \( \frac{16}{40} \) = \( \frac{2}{5} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 40 cards left, which gives 16.

(t) not a natural number

Answer:
1. Favourable cards: 8 out of 40 cards left in the deck.
2. P = \( \frac{8}{40} \) = \( \frac{1}{5} \)

Teacher's Note:
a) An ace counts as 1 and number cards from 2 to 10 are natural numbers, but jack, queen and king are not natural numbers.
b) So the favourable cards are the face cards left in the deck, which number 8 out of 40.

7. Five cards - the ten, jack, queen, king and ace of diamonds, are removed from the well-shuffled 52 playing cards. One card is then picked up at random. Find the probability of getting

(a) neither a heart nor a king

Answer:
1. Favourable cards: 32 out of 47 cards left in the deck.
2. P = \( \frac{32}{47} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 32 favourable cards out of 47, so the probability is \( \frac{32}{47} \).

(b) neither an ace nor a king

Answer:
1. Favourable cards: 41 out of 47 cards left in the deck.
2. P = \( \frac{41}{47} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 41 favourable cards out of 47, so the probability is \( \frac{41}{47} \).

(c) neither a red card nor a queen card

Answer:
1. Favourable cards: 24 out of 47 cards left in the deck.
2. P = \( \frac{24}{47} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 24 favourable cards out of 47, so the probability is \( \frac{24}{47} \).

(d) a black card or an ace

Answer:
1. Favourable cards: 27 out of 47 cards left in the deck.
2. P = \( \frac{27}{47} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 27 favourable cards in total out of 47 cards left.

(e) either a heart or a spade card

Answer:
1. Favourable cards: 26 out of 47 cards left in the deck.
2. P = \( \frac{26}{47} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 26 favourable cards in total out of 47 cards left.

(f) a king card

Answer:
1. Favourable cards: 3 out of 47 cards left in the deck.
2. P = \( \frac{3}{47} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(g) a heart card

Answer:
1. Favourable cards: 13 out of 47 cards left in the deck.
2. P = \( \frac{13}{47} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(h) a red card

Answer:
1. Favourable cards: 21 out of 47 cards left in the deck.
2. P = \( \frac{21}{47} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(i) a black card

Answer:
1. Favourable cards: 26 out of 47 cards left in the deck.
2. P = \( \frac{26}{47} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(j) a spade card

Answer:
1. Favourable cards: 13 out of 47 cards left in the deck.
2. P = \( \frac{13}{47} \)

Teacher's Note:
a) The denominator is the number of cards left in the deck after any removal.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(k) a diamond card

Answer:
1. Favourable cards: 8 out of 47 cards left in the deck.
2. P = \( \frac{8}{47} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(l) a club card

Answer:
1. Favourable cards: 13 out of 47 cards left in the deck.
2. P = \( \frac{13}{47} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(m) either an ace card or black card

Answer:
1. Favourable cards: 27 out of 47 cards left in the deck.
2. P = \( \frac{27}{47} \)

Teacher's Note:
a) Add the two groups and subtract the cards that belong to both, so no card is counted twice.
b) Forgetting the overlap is the usual error for an 'or' question.
c) There are 27 favourable cards in total out of 47 cards left.

(n) an ace card

Answer:
1. Favourable cards: 3 out of 47 cards left in the deck.
2. P = \( \frac{3}{47} \)

Teacher's Note:
a) Write the favourable count over the deck size, then reduce the fraction to lowest terms.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(o) a face card

Answer:
1. Favourable cards: 9 out of 47 cards left in the deck.
2. P = \( \frac{9}{47} \)

Teacher's Note:
a) Count the favourable cards, then divide by the total number of cards still in the deck.
b) The deck has 47 cards after removal, so the denominator is 47, not 52.

(p) a face card with red colour

Answer:
1. Favourable cards: 3 out of 47 cards left in the deck.
2. P = \( \frac{3}{47} \)

Teacher's Note:
a) Count only the cards that satisfy both conditions together.
b) Do not add the two groups for an 'and' question; that is the 'or' rule.

(q) neither '10' card nor an ace

Answer:
1. Favourable cards: 41 out of 47 cards left in the deck.
2. P = \( \frac{41}{47} \)

Teacher's Note:
a) Count the cards that avoid both features, which is the complement of the cards with either feature.
b) Subtracting only one of the two groups leaves cards from the other group in the count, which is a common mistake.
c) This gives 41 favourable cards out of 47, so the probability is \( \frac{41}{47} \).

(r) an even number card

Answer:
1. Favourable cards: 19 out of 47 cards left in the deck.
2. P = \( \frac{19}{47} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 47 cards left, which gives 19.

(s) an odd number card

Answer:
1. Favourable cards: 16 out of 47 cards left in the deck.
2. P = \( \frac{16}{47} \)

Teacher's Note:
a) Even cards are 2, 4, 6, 8 and 10, and odd cards are 3, 5, 7 and 9. The ace and the face cards are neither even nor odd, so they are not counted.
b) Count the matching cards among the 47 cards left, which gives 16.

(t) not a natural number

Answer:
1. Favourable cards: 9 out of 47 cards left in the deck.
2. P = \( \frac{9}{47} \)

Teacher's Note:
a) An ace counts as 1 and number cards from 2 to 10 are natural numbers, but jack, queen and king are not natural numbers.
b) So the favourable cards are the face cards left in the deck, which number 9 out of 47.

IMPORTANT TIPS

• Coin. A coin has two faces termed as Head and Tail.
• Dice. A dice is a small cube which has between one to six spots or numbers on its sides, which is used in board games like Ludo, Snakes & Ladders. In old fashioned English, dice was used only as a plural form, but now dice is used as both the singular and plural form.
• Cards. A pack of playing cards consists of four suits called Hearts, Spades, Diamonds and Clubs. Each suit consists of 13 cards. Each suit consists of one ace, three face cards and nine number cards.

LEARNING OBJECTIVES

At the end of this lesson you will be able to

• define the sample space accurately.
• find that the total number of favourable outcomes are those outcomes in the sample space that are favourable to the occurence of an event.
• find the probability of an event.

Very Short Answer 

Question. In a family of 3 children calculate the probability of having at least one boy.
Answer: Sol. Let ‘b’ denote a boy and ‘g’ denote a girl. Now, the following outcomes may be noted.
S = {bbb, bbg, bgb, bgg, gbg, ggb, gbb, ggg}
\( \therefore \) Total outcomes = 8
No. of outcomes with atleast one boy = 7
\( \therefore \) Probability of having atleast one boy = \( \frac{7}{8} \)

Question. A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
Answer: Sol. Total English alphabets = 26
Number of consonants = 21
\( \therefore \) P (letter is a consonant) = \( \frac{21}{26} \)

Question. A box contains cards numbered 6 to 50. A card is drawn at random from the box. Calculate the probability that the drawn card has a number which is a perfect square.
Answer: Sol. Total number of cards = 50 – 6 + 1 = 45
Perfect square numbers are 9, 16, 25, 36, 49, i.e., 5 numbers
\( \therefore \) P(a prefect square) = \( \frac{5}{45} = \frac{1}{9} \)

Question. Cards marked with number 3, 4, 5, ...., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
Answer: Sol. Total no. of cards = 50 – 3 + 1 = 48
Perfect square number cards are 4, 9, 16, 25, 36, 49 i.e., 6 cards
\( \therefore \) P(perfect square number) = \( \frac{6}{48} = \frac{1}{8} \)

Question. A number is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1?
Answer: Sol. \( (-3)^2 = 9 \); \( (-2)^2 = 4 \); \( (-1)^2 = 1 \); \( (0)^2 = 0 \); \( (1)^2 = 1 \); \( (2)^2 = 4 \); \( (3)^2 = 9 \)
\( \therefore \) P(Sq. of nos. \( \leq \) 1) = \( \frac{3}{7} \)

Question. If two different dice are rolled together, calculate the probability of getting an even number on both dice.
Answer: Sol. Two dice can be thrown as 6 \( \times \) 6 = 36 ways. Even numbers on both the dice can be obtained as {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}, i.e., 9 ways.
\( \therefore \) P(even numbers) = \( \frac{9}{36} = \frac{1}{4} \)

Question. Two different dices are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.
Answer: Sol. Total outcomes = \( 6^n = 6^2 = 36 \)
Possible outcomes having the product of the two numbers on the top of the dice as 6 are (3 \( \times \) 2, 2 \( \times \) 3, 6 \( \times \) 1, 1 \( \times \) 6), i.e, 4.
P(Product of two nos. is 6) = \( \frac{4}{36} = \frac{1}{9} \)

Question. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?
Answer: Sol. P(rotten apples) = \( \frac{\text{No. of rotten apples}}{\text{Total apples}} \)
\( 0.18 = \frac{\text{No. of rotten apples}}{900} \)
\( \therefore \) No. of rotten apples = 900 \( \times \) 0.18 = 162

Short Answer-I 

Question. A coin is tossed two times. Find the probability of getting both heads or both tails.
Answer: Sol. S = {HH, HT, TH, TT} = 4
P (both heads or both tails) = P (both heads) + P (both tails) = \( \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)

Question. In a simultaneous toss of two coins, find the probability of getting: (i) exactly one head, (ii) atmost one head.
Answer: Sol. The sample space is given by S = {HH, HT, TH, TT}. Total events n(S) = 4.
(i) exactly one head = {HT, TH} = 2. P(exactly one head) = \( \frac{2}{4} = \frac{1}{2} \).
(ii) atmost one head = {HT, TH, TT} = 3. P(atmost one head) = \( \frac{3}{4} \).

Question. Two coins are tossed simultaneously. Find the probability of getting at least one head.
Answer: Sol. S = {HH, HT, TH, TT}, i.e., 4.
\( \therefore \) P (atleast one head) = \( \frac{3}{4} \)

Question. Three coins are tossed simultaneously. Find the probability of getting exactly two heads.
Answer: Sol. S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8.
P(exactly two heads) = \( \frac{3}{8} \)

Question. Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.
Answer: Sol. S = {HH, HT, TH, TT}, i.e., 3.
P(at least one tail) = \( \frac{3}{4} \)

Question. Two dice are thrown simultaneously. Find the probability of getting a doublet.
Answer: Sol. Two dice can be thrown as 6 \( \times \) 6 = 36 ways. “a doublet” can be obtained by (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), i.e., 6 ways.
P(a doublet) = \( \frac{6}{36} = \frac{1}{6} \)

Question. A dice is tossed once. Find the probability of getting an even number or a multiple of 3.
Answer: Sol. S = {1, 2, 3, 4, 5, 6} = 6. ‘an even number or multiple of 3’ are 2, 3, 4, 6, i.e., 4 numbers.
\( \therefore \) Required probability = \( \frac{4}{6} = \frac{2}{3} \)

Question. Two different dice are tossed together. Find the probability (i) that the number on each die is even. (ii) that the sum of numbers appearing on the two dice is 5.
Answer: Sol. Two dice can be thrown as 6 \( \times \) 6 = 36 ways.
(i) The probability of number on each die is even are {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}, i.e., 9 ways. \( \therefore \) P(no. on each die is even) = \( \frac{9}{36} = \frac{1}{4} \).
(ii) “Sum is 5” can be obtained as (2, 3), (3, 2), (1, 4), (4, 1), i.e., 4 ways. \( \therefore \) P(Sum 5) = \( \frac{4}{36} = \frac{1}{9} \)

Short Answer-II 

Question. A coin is tossed two times. Find the probability of getting at least one head.
Answer: Sol. S = {HH, HT, TH, TT}. Total number of ways = 4. Atleast one head = {HH, HT, TH}, i.e., 3 ways. \( \therefore \) P (atleast one head) = \( \frac{3}{4} \)

Question. A coin is tossed two times. Find the probability of getting not more than one head.
Answer: Sol. S = {HH, HT, TH, TT} = 4. Favourable cases are HT, TH, TT i.e., 3 cases. \( \therefore \) P (not more than 1 head) = \( \frac{3}{4} \)

Question. Three distinct coins are tossed together. Find the probability of getting (i) at least 2 heads (ii) at most 2 heads.
Answer: Sol. Total number of possible outcomes = \( 2^n = 2^3 = 8 \) (HHH, TTT, HHT, THH, THT, HTH, TTH, HTT).
(i) Possible outcomes of at least two heads = 4 (HHT, THH, HHH, HTH). \( \therefore \) P(at least two heads) = \( \frac{4}{8} = \frac{1}{2} \).
(ii) Possible outcomes of at most two heads = 7 (HHT, TTT, THH, THT, HTH, TTH, HTT). \( \therefore \) P(at most two heads) = \( \frac{7}{8} \)

Question. Three different coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at least two tails.
Answer: Sol. S {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. Total number of ways = 8.
(i) Exactly two heads = HHT, HTH, THH, i.e., 3 ways. \( \therefore \) P (exactly two heads) = \( \frac{3}{8} \).
(ii) Atleast two heads = HHT, HTH, THH, HHH i.e., 4 ways. \( \therefore \) P (atleast two heads) = \( \frac{4}{8} = \frac{1}{2} \).
(iii) Atleast two tails = TTH, THT, HTT, TTT i.e., 4 ways. \( \therefore \) P (atleast two tails) = \( \frac{4}{8} = \frac{1}{2} \)

Question. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5.
Answer: Sol. Total number of tickets = 40. ‘A multiple of 5’ are 5, 10, 15, 20, … 40, i.e., 8 tickets. \( \therefore \) P (A multiple of 5) = \( \frac{8}{40} = \frac{1}{5} \)

Question. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.
Answer: Sol. Multiples of 3 & 4 are 12, 24, 36, 48, i.e., 4 nos. P(a multiple of 3 and 4) = \( \frac{4}{50} = \frac{2}{25} \)

Question. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is 1/4. The probability of selecting a blue ball at random from the same jar is 1/3. If the jar contains 10 orange balls, find the total number of balls in the jar.
Answer: Sol. P(Red) = \( \frac{1}{4} \), P (blue) = \( \frac{1}{3} \). As we know, Total Probability = 1.
\( \Rightarrow \) P(orange) = \( 1 - \frac{1}{4} - \frac{1}{3} = \frac{5}{12} \).
\( \Rightarrow \text{P(orange)} = \frac{\text{Total no. of orange balls}}{\text{Total no. of balls}} \).
\( \Rightarrow \frac{5}{12} = \frac{10}{\text{Total no. of balls}} \). \( \therefore \) Total no. of balls = \( \frac{10 \times 12}{5} = 24 \).

Question. A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is 3/10 and that of a black ball is 2/5, then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.
Answer: Sol. Let W, B and R denote the White, Black and Red balls respectively. We know, Total probability = 1.
P(W) = \( \frac{3}{10} \), P(B) = \( \frac{2}{5} \). \( \therefore \) P(R) = \( 1 - \frac{3}{10} - \frac{2}{5} = \frac{3}{10} \).
P(B) \( \times \) Total no. of balls = No. of Black balls. \( \frac{2}{5} \times \text{(Total number of balls)} = 20 \).
\( \therefore \) Total number of balls = \( \frac{20 \times 5}{2} = 50 \).

Question. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.
Answer: Sol. Let the number of black balls = x. The number of white balls = 15. \( \therefore \) Total number of balls = x + 15.
\( \therefore \) P(black ball) = 3P(White balls). \( \frac{x}{x + 15} = 3 \left( \frac{15}{x + 15} \right) \). \( \frac{x}{x + 15} = \frac{45}{x + 15} \).
\( \because x + 15 \neq 0, \therefore x = 45 \). Hence, the number of black balls = 45.

Question. In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11?
Answer: Sol. Two dice can be thrown in 6 \( \times \) 6 = 36 ways.
(i) “a prime number on each dice” can be obtained as (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5), i.e., 9 ways.
\( \therefore \) P(a prime no. on each dice) = \( \frac{9}{36} = \frac{1}{4} \).
(ii) “a total of 9 or 11” can be obtained as (3,6),(6,3),(4,5),(5,4) [Total '9'] and (5,6),(6,5) [Total '11'], i.e., 6 ways.
\( \therefore \) P(a total of 9 or 11) = \( \frac{6}{36} = \frac{1}{6} \)

Question. A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal’ another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that: (i) Ramesh will buy the selected shirt? (ii) ‘Kewal’ will buy the selected shirt?
Answer: Sol. Total = 88 + 8 + 4 = 100. No. of good shirts = 88.
(i) P(Ramesh buys a shirt) = P(good shirts) = \( \frac{88}{100} = \frac{22}{25} \).
(ii) No. of shirts without major defect = 96. P(Kewal buys a shirt) = P(shirts without major defect) = \( \frac{88 + 8}{100} = \frac{96}{100} = \frac{24}{25} \)

Question. All red face cards are removed from a pack of playing cards. The remaining cards were well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is (i) a red card (ii) a face card (iii) a card of clubs.
Answer: Sol. Number of red face cards removed = 6. \( \therefore \) Remaining cards = 52 – 6 = 46. Hence, Total no. of outcomes = 46.
(i) Possible outcomes of red cards = 26 – 6 = 20. \( \therefore \) P(a red card) = \( \frac{20}{46} = \frac{10}{23} \).
(ii) Possible outcomes of face cards = 6. \( \therefore \) P(a face card) = \( \frac{6}{46} = \frac{3}{23} \).
(iii) Possible outcomes of card of clubs = 13. \( \therefore \) P(a card of clubs) = \( \frac{13}{46} \)

Question. From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is: (i) a black King (ii) a card of red colour (iii) a card of black colour.
Answer: Sol. Total cards in the pack = 52. Cards removed = 2(Jacks) + 2(Queens) + 2(Kings) = 6. \( \therefore \) Remaining cards = 52 – 6 = 46.
(i) Number of black Kings = 2. \( \therefore \) P(a black King) = \( \frac{2}{46} = \frac{1}{23} \).
(ii) Total red cards in the pack = 26. Red cards removed = 6. Remaining red cards = 26 – 6 = 20. \( \therefore \) P(a card of red colour) = \( \frac{20}{46} = \frac{10}{23} \).
(iii) Total black cards in the pack = 26. \( \therefore \) P(a card of black colour) = \( \frac{26}{46} = \frac{13}{23} \)

Question. There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the prob ability that the number on the selected card (i) is divisible by 9 and is a perfect square (ii) is a prime number greater than 80.
Answer: Sol. Total cards = 100.
(i) Numbers which are “Divisible by 9 and perfect squares” are 9, 36, 81, i.e., 3. \( \therefore \) P(Divisible by 9 & perfect square) = \( \frac{3}{100} \).
(ii) Numbers which are “prime numbers greater than 80” are 83, 89, 97, i.e., 3. \( \therefore \) P(Prime nos. > 80) = \( \frac{3}{100} \)

Question. Two different dice are rolled together. Find the probability of getting: (i) the sum of numbers on two dice to be 5. (ii) even numbers on both dice.
Answer: Sol. Total possible outcomes = \( 6^n = 6^2 = 36 \).
(i) The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) when the sum of numbers on two dice is 5, i.e., 4. \( \therefore \) Required Probability, P(E) = \( \frac{4}{36} = \frac{1}{9} \).
(ii) The possible outcomes are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) for even numbers on both dice; 9. \( \therefore \) Required Probability, P(E) = \( \frac{9}{36} = \frac{1}{4} \)

Question. Two different dice are thrown together. Find the probability of: (i) getting a number greater than 3 on each die (ii) getting a total of 6 or 7 of the numbers on two dice.
Answer: Sol. Two dice can be thrown in 6 \( \times \) 6 = 36 ways.
(i) “getting a number > 3 on each die” can be obtained as (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), i.e., 9 ways. P(number > 3 on each die) = \( \frac{9}{36} = \frac{1}{4} \).
(ii) “a total of 6 or 7 can be obtained as (2, 4),(4,2),(3,3),(1,5),(5,1) [Total '6'] and (1,6),(6,1),(2,5),(5,2),(3,4),(4,3) [Total '7'], i.e., 11 ways. P(a total of 6 or 7) = \( \frac{11}{36} \)

Question. Two different dice are thrown together. Find the probability that the numbers obtained. (i) have a sum less than 6 (ii) have a product less than 16 (iii) is a doublet of odd numbers.
Answer: Sol. Two dice can be thrown in (6 \( \times \) 6) = 36 ways.
(i) Sum less than ‘6’ are (1, 1) (1, 2) (1, 3) (1, 4), (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (4, 1) i.e., 10 ways. \( \therefore \) P(sum < 6) = \( \frac{10}{36} = \frac{5}{18} \).
(ii) Product less than ‘16’ are (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (4, 1) (4, 2) (4, 3) (5, 1) (5, 2) (5, 3) (6, 1) (6, 2) i.e., 25 ways. \( \therefore \) P(product less than 16) = \( \frac{25}{36} \).
(iii) Doublet of odd numbers: (1, 1), (3, 3), (5, 5) i.e., 3 ways. \( \therefore \) P(doublet of odd nos.) = \( \frac{3}{36} = \frac{1}{12} \)