CBSE Class 10 Mathematics Probability Assignment Set 15

Probability of an Event

Question 1. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds.
Answer:
Total number of possible outcomes = 52.
(i) Red kings: king of hearts and king of diamonds = 2. \( P = \frac{2}{52} = \frac{1}{26} \).
(ii) Face cards (kings, queens, jacks) = 12. \( P = \frac{12}{52} = \frac{3}{13} \).
(iii) Red face cards = 3 in hearts + 3 in diamonds = 6. \( P = \frac{6}{52} = \frac{3}{26} \).
(iv) There is only one jack of hearts. \( P = \frac{1}{52} \).
(v) Spades = 13. \( P = \frac{13}{52} = \frac{1}{4} \).
(vi) There is one queen of diamonds. \( P = \frac{1}{52} \).
In simple words: Know the composition of a pack: 4 suits of 13 cards, 12 face cards, 26 red 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 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?
Answer:
(i) There is 1 queen among 5 cards: \( P = \frac{1}{5} \).
(ii) After the queen is put aside, 4 cards remain: ten, jack, king and ace.
(a) \( P(\text{ace}) = \frac{1}{4} \).
(b) There is no queen left, so \( P(\text{queen}) = \frac{0}{4} = 0 \) (an impossible event).
In simple words: After a card is removed, the total number of cards drops by one.

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. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer:
Number of good pens = 132 and total pens = 132 + 12 = 144.
\( P(\text{good pen}) = \frac{132}{144} = \frac{11}{12} \).
In simple words: Divide the number of good pens by the total number of pens.

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. (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer:
(i) \( P(\text{defective}) = \frac{4}{20} = \frac{1}{5} \).
(ii) The first bulb was good and is not replaced, so 19 bulbs are left, of which 4 are defective and \( 19 - 4 = 15 \) are good.
\( P(\text{not defective}) = \frac{15}{19} \).
In simple words: Without replacement, the total goes down by one after the first 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 5. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Answer:
Total number of discs = 90.
(i) Single-digit numbers are 1 to 9, so two-digit numbers = 90 - 9 = 81. \( P = \frac{81}{90} = \frac{9}{10} \).
(ii) Perfect squares up to 90: 1, 4, 9, 16, 25, 36, 49, 64, 81 (9 numbers). \( P = \frac{9}{90} = \frac{1}{10} \).
(iii) Multiples of 5: 5, 10, ..., 90, that is 18 numbers. \( P = \frac{18}{90} = \frac{1}{5} \).
In simple words: Count the favourable numbers carefully and divide by 90.

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. A child has a die whose six faces show the letters as given below: ABCDEA
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Answer:
Total number of faces = 6.
(i) The letter A appears on 2 faces. \( P(A) = \frac{2}{6} = \frac{1}{3} \).
(ii) The letter D appears on 1 face. \( P(D) = \frac{1}{6} \).
In simple words: Count how many faces carry the required letter.

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. Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m?

Answer:
The area of the rectangle (all possible outcomes) \( = 3 \times 2 = 6 \) m\( ^2 \).
The circle has diameter 1 m, so radius \( = \frac{1}{2} \) m and its area \( = \pi \left( \frac{1}{2} \right)^2 = \frac{\pi}{4} \) m\( ^2 \).
\( P = \frac{\text{area of circle}}{\text{area of rectangle}} = \frac{\pi/4}{6} = \frac{\pi}{24} \).
In simple words: Here probability is the ratio of the favourable area to the total area.

Teacher's Note:
1. When the outcomes are spread over a region, compare areas instead of counting.
2. Convert the diameter to the radius before finding the area of the circle.

Question 8. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) she will buy it? (ii) she will not buy it?
Answer:
Total pens = 144; defective = 20; good = 124.
(i) \( P(\text{buys}) = P(\text{good}) = \frac{124}{144} = \frac{31}{36} \).
(ii) \( P(\text{does not buy}) = 1 - \frac{31}{36} = \frac{5}{36} \).
In simple words: She buys only a good pen, so her buying is the event 'good pen'.

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.

Teacher's Note:
1. Make a 6 by 6 table of sums to count each total.
2. The 11 possible sums are not equally likely outcomes, so 1/11 is wrong.

Question 10. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Answer:
The 8 equally likely outcomes are HHH, HHT, HTH, THH, TTH, THT, HTT, TTT.
Hanif wins in HHH and TTT (2 outcomes), so he loses in the other 6 outcomes.
\( P(\text{loses}) = \frac{6}{8} = \frac{3}{4} \).
In simple words: Hanif loses unless the three tosses are all alike.

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 die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
Answer:
Total number of outcomes \( = 6 \times 6 = 36 \).
(i) \( P(\text{5 does not come up either time}) = 1 - \frac{11}{36} = \frac{25}{36} \).
(ii) Outcomes in which 5 comes up at least once: (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5), which is 11 outcomes. \( P = \frac{11}{36} \).
In simple words: 'Not 5 either time' is the complement of 'at least one 5'.

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. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes: two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \( \frac{1}{3} \).
(ii) If a die is thrown, there are two possible outcomes: an odd number or an even number. Therefore, the probability of getting an odd number is \( \frac{1}{2} \).
Answer:
(i) Not correct. When two coins are tossed there are 4 equally likely outcomes: HH, HT, TH and TT. So \( P(\text{two heads}) = \frac{1}{4} \), \( P(\text{two tails}) = \frac{1}{4} \) and \( P(\text{one of each}) = \frac{2}{4} = \frac{1}{2} \). The three outcomes mentioned are not equally likely.
(ii) Correct. The outcomes 1, 2, 3, 4, 5, 6 contain 3 odd and 3 even numbers, so \( P(\text{odd}) = \frac{3}{6} = \frac{1}{2} \). Here the two outcomes (odd, even) are equally likely.
In simple words: Probability can be found as 1 over the number of outcomes only when the outcomes are equally likely.

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.