CBSE Class 10 Mathematics Probability Assignment Set 16

Very Short Answer Type Questions

Question 1. A card is selected from a deck of 52 cards. Find the probability that the card drawn is a black face card.
Answer:
Black face cards: king, queen and jack of spades and of clubs = 6 cards.
\( P = \frac{6}{52} = \frac{3}{26} \).
In simple words: There are 3 face cards in each of the two black suits.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 2. Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 3.
Answer:
Total outcomes = 36. Pairs with difference 3: (1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3), that is 6 outcomes.
\( P = \frac{6}{36} = \frac{1}{6} \).
In simple words: List the pairs whose numbers differ by 3 in either order.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 3. A letter of the English alphabet is chosen at random. Determine the probability that the letter is a consonant.
Answer:
There are 26 letters and 5 vowels (A, E, I, O, U), so 21 consonants.
\( P = \frac{21}{26} \).
In simple words: Subtract the 5 vowels from 26.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 4. There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs. 100 each, 100 of them contain a cash prize of Rs. 50 each and 200 of them contain a cash prize of Rs. 10 each and the rest do not contain any cash prize. If they are well shuffled and an envelope is picked, what is the probability that it contains no cash prize?
Answer:
Envelopes with a prize = 10 + 100 + 200 = 310, so envelopes with no prize = 1000 - 310 = 690.
\( P(\text{no prize}) = \frac{690}{1000} = \frac{69}{100} \).
In simple words: Count the envelopes with a prize and subtract from 1000.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 5. Two dice are thrown at the same time and the product of the numbers appearing on them is noted. Find the probability that the product is less than 9.
Answer:
Total outcomes = 36. Favourable outcomes (product less than 9): first number 1: all 6; first number 2: second 1, 2, 3, 4 (4 outcomes); first number 3: 1, 2 (2); first number 4: 1, 2 (2); first number 5: 1 (1); first number 6: 1 (1).
Total favourable = 6 + 4 + 2 + 2 + 1 + 1 = 16. \( P = \frac{16}{36} = \frac{4}{9} \).
In simple words: Go through the first number one by one and count the second numbers that keep the product below 9.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 6. Two dice are thrown simultaneously at random. Find the probability of getting a sum of eight.
Answer:
Sum 8: (2, 6), (6, 2), (3, 5), (5, 3), (4, 4), that is 5 outcomes out of 36.
\( P = \frac{5}{36} \).
In simple words: List the pairs adding to 8.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 7. A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.
Answer:
Total outcomes = 8. Success: HHH and TTT (2 outcomes).
\( P(\text{success}) = \frac{2}{8} = \frac{1}{4} \), so \( P(\text{losing}) = 1 - \frac{1}{4} = \frac{3}{4} \).
In simple words: Losing is the complement of success.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 8. A die is thrown once. Find the probability of getting a number which lies between 2 and 6.
Answer:
Numbers between 2 and 6: 3, 4, 5 (3 numbers out of 6).
\( P = \frac{3}{6} = \frac{1}{2} \).
In simple words: The numbers 2 and 6 themselves are not included.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 9. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.
Answer:
Let there be x blue balls. Total = x + 5. \( P(\text{blue}) = \frac{x}{x + 5} \) and \( P(\text{red}) = \frac{5}{x + 5} \).
\( \frac{x}{x + 5} = 3 \times \frac{5}{x + 5} \Rightarrow x = 15 \). There are 15 blue balls.
In simple words: Equate P(blue) to three times P(red); both have the same denominator.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Short Answer Type Questions

Question 10. Five cards, the two, jack, queen, king and ace of diamonds, are well shuffled with their faces downward. One card is then picked up at random.
(i) What is the probability that the card is the king?
(ii) If the king is drawn and put aside, what is the probability that the second card picked up is (a) a face card? (b) a king?
Answer:
(i) \( P(\text{king}) = \frac{1}{5} \).
(ii) After the king is put aside, 4 cards remain: two, jack, queen, ace.
(a) Face cards left: jack and queen = 2. \( P = \frac{2}{4} = \frac{1}{2} \).
(b) No king is left. \( P = \frac{0}{4} = 0 \).
In simple words: Remove the drawn card before counting for the second draw.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 11. A bag contains 30 balls of which x are red, 2x are yellow and 3x are blue. A ball is selected at random. What is the probability that it is (i) not red? (ii) yellow?
Answer:
\( x + 2x + 3x = 30 \Rightarrow 6x = 30 \Rightarrow x = 5 \). So 5 red, 10 yellow, 15 blue.
(i) \( P(\text{not red}) = 1 - \frac{5}{30} = \frac{25}{30} = \frac{5}{6} \).
(ii) \( P(\text{yellow}) = \frac{10}{30} = \frac{1}{3} \).
In simple words: Find x from the total number of balls first.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 12. A bag contains 12 green cards and some blue cards. If the probability of drawing a green card is double that of a blue card, determine the number of blue cards in the bag.
Answer:
Let there be b blue cards. Total = 12 + b.
\( \frac{12}{12 + b} = 2 \times \frac{b}{12 + b} \Rightarrow 12 = 2b \Rightarrow b = 6 \). There are 6 blue cards.
In simple words: The denominators are equal, so the numbers of cards are in the same ratio as the probabilities.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 13. A selection committee interviewed some people for the post of CEO. So, they called male and female candidates in the ratio 5 : 6. What is the probability of a male candidate being selected?
Answer:
Let the numbers of male and female candidates be 5k and 6k. Total = 11k.
\( P(\text{male}) = \frac{5k}{11k} = \frac{5}{11} \).
In simple words: A ratio 5 : 6 means 5 male out of every 11 candidates.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 14. The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is (i) a heart (ii) a king.
Answer:
Remaining cards = 52 - 3 = 49.
(i) Hearts are untouched: 13. \( P = \frac{13}{49} \).
(ii) Kings left: 4 - 1 = 3. \( P = \frac{3}{49} \).
In simple words: Update the total to 49 and recount the required cards.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 15. 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:
(i) Ramesh buys only good shirts: \( P = \frac{88}{100} = \frac{22}{25} \).
(ii) Kewal buys good shirts and shirts with minor defects: \( 88 + 8 = 96 \). \( P = \frac{96}{100} = \frac{24}{25} \).
In simple words: Decide which shirts each shopkeeper accepts, then count them.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 16. 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 the drawn card is (i) a black king (ii) a card of red colour (iii) a card of black colour.
Answer:
Cards removed = 3 + 3 = 6. Remaining = 46.
(i) Black kings: 2. \( P = \frac{2}{46} = \frac{1}{23} \).
(ii) Red cards left: 26 - 6 = 20. \( P = \frac{20}{46} = \frac{10}{23} \).
(iii) Black cards: 26. \( P = \frac{26}{46} = \frac{13}{23} \).
In simple words: After removing 6 red cards the total is 46.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 17. 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:
Let there be x black balls. \( \frac{x}{x + 15} = 3 \times \frac{15}{x + 15} \Rightarrow x = 45 \). There are 45 black balls.
In simple words: The two probabilities share the same denominator, so the numbers are in the ratio 3 : 1.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 18. A box contains cards numbered 1 to 90. A card is drawn at random from the box. Find the probability that the selected card is (i) a two-digit number (ii) a perfect square number.
Answer:
Total cards = 90.
(i) Two-digit numbers: 10 to 90 = 81. \( P = \frac{81}{90} = \frac{9}{10} \).
(ii) Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9. \( P = \frac{9}{90} = \frac{1}{10} \).
In simple words: Count carefully the two-digit numbers and the squares.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.