Exercise 13.2 Solutions – Class X Mathematics

Exercise 13.2 Solutions

Probability – Class X Mathematics

Question 1

A bag contains 3 red balls and 5 black balls. A ball is selected at random from the bag. What is the probability that the ball selected is:

(i) red? (ii) not red?

Solution:

Total number of balls = 3 (red) + 5 (black) = 8

(i) Probability of selecting a red ball = \(\frac{\text{Number of red balls}}{\text{Total balls}} = \frac{3}{8}\)

(ii) Probability of not selecting a red ball = 1 – Probability of selecting red ball = \(1 – \frac{3}{8} = \frac{5}{8}\)

Question 2

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be:

(i) red? (ii) white? (iii) not green?

Solution:

Total marbles = 5 (red) + 8 (white) + 4 (green) = 17

(i) Probability of red marble = \(\frac{5}{17}\)

(ii) Probability of white marble = \(\frac{8}{17}\)

(iii) Probability of not green marble = 1 – Probability of green marble = \(1 – \frac{4}{17} = \frac{13}{17}\)

Question 3

A Kiddy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin:

(i) will be a 50p coin? (ii) will not be a ₹5 coin?

Solution:

Total coins = 100 (50p) + 50 (₹1) + 20 (₹2) + 10 (₹5) = 180

(i) Probability of 50p coin = \(\frac{100}{180} = \frac{5}{9}\)

(ii) Probability of not ₹5 coin = 1 – Probability of ₹5 coin = \(1 – \frac{10}{180} = \frac{170}{180} = \frac{17}{18}\)

Question 4

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Diagram Description: A rectangular fish tank containing 5 male fish (shown as smaller, colorful fish) and 8 female fish (shown as larger, less colorful fish). The shopkeeper is using a net to catch one fish at random.

Solution:

Total fish = 5 (male) + 8 (female) = 13

Probability of male fish = \(\frac{5}{13}\)

Question 5

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See figure), and these are equally likely outcomes. What is the probability that it will point at:

(i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

Diagram Description: A circular spinner divided into 8 equal sectors numbered 1 through 8 clockwise. An arrow is fixed at the center that can spin freely.

Solution:

Total possible outcomes = 8

(i) Probability of pointing at 8 = \(\frac{1}{8}\)

(ii) Odd numbers = {1, 3, 5, 7} → 4 outcomes. Probability = \(\frac{4}{8} = \frac{1}{2}\)

(iii) Numbers > 2 = {3, 4, 5, 6, 7, 8} → 6 outcomes. Probability = \(\frac{6}{8} = \frac{3}{4}\)

(iv) Numbers < 9 = {1, 2, 3, 4, 5, 6, 7, 8} → 8 outcomes. Probability = \(\frac{8}{8} = 1\)

Question 6

One card is selected 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

Solution:

Total cards = 52

(i) King of red colour: There are 2 (King of Hearts and King of Diamonds). Probability = \(\frac{2}{52} = \frac{1}{26}\)

(ii) Face cards: Jack, Queen, King in each suit → 3 × 4 = 12. Probability = \(\frac{12}{52} = \frac{3}{13}\)

(iii) Red face cards: Face cards in hearts and diamonds → 3 × 2 = 6. Probability = \(\frac{6}{52} = \frac{3}{26}\)

(iv) Jack of hearts: Only 1 card. Probability = \(\frac{1}{52}\)

(v) Spade cards: 13. Probability = \(\frac{13}{52} = \frac{1}{4}\)

(vi) Queen of diamonds: Only 1 card. Probability = \(\frac{1}{52}\)

Question 7

Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is selected at random.

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

(ii) If the queen is selected and put aside (without replacement), what is the probability that the second card selected is (a) an ace? (b) a queen?

Solution:

Total cards initially = 5

(i) Probability of queen = \(\frac{1}{5}\)

(ii) After removing queen, remaining cards = 4

(a) Probability of ace = \(\frac{1}{4}\) (only ace of diamonds left)

(b) Probability of queen = \(\frac{0}{4} = 0\) (queen has been removed)

Question 8

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.

Solution:

Total pens = 12 (defective) + 132 (good) = 144

Probability of good pen = \(\frac{132}{144} = \frac{11}{12}\)

Question 9

A lot of 20 bulbs contain 4 defective ones. One bulb is selected at random from the lot. What is the probability that this bulb is defective? Suppose the bulb selected in previous case is not defective and is not replaced. Now one bulb is selected at random from the rest. What is the probability that this bulb is not defective?

Solution:

First selection:

Total bulbs = 20, Defective = 4

Probability of defective bulb = \(\frac{4}{20} = \frac{1}{5}\)

Second selection (after removing one good bulb):

Remaining bulbs = 19, Non-defective = 16 – 1 = 15

Probability of not defective bulb = \(\frac{15}{19}\)

Question 10

A box contains 90 discs which are numbered from 1 to 90. If one disc is selected 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.

Solution:

Total discs = 90

(i) Two-digit numbers: 10 to 90 → 81 numbers. Probability = \(\frac{81}{90} = \frac{9}{10}\)

(ii) Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 → 9 numbers. Probability = \(\frac{9}{90} = \frac{1}{10}\)

(iii) Divisible by 5: 5, 10, 15, …, 90 → 18 numbers. Probability = \(\frac{18}{90} = \frac{1}{5}\)

Question 11

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

Diagram Description: A rectangular region with length 3m and width 2m. Inside it, there’s a circle with diameter 1m (radius 0.5m) centered in the rectangle.

Solution:

Area of rectangle = length × width = 3 × 2 = 6 m²

Area of circle = πr² = π(0.5)² = 0.25π m²

Probability = \(\frac{\text{Area of circle}}{\text{Area of rectangle}} = \frac{0.25π}{6} = \frac{π}{24}\)

Question 12

A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that:

(i) She will buy it? (ii) She will not buy it?

Solution:

Assuming Sudha will buy if the pen is good:

Good pens = 144 – 20 = 124

(i) Probability she will buy = Probability of good pen = \(\frac{124}{144} = \frac{31}{36}\)

(ii) Probability she will not buy = Probability of defective pen = \(\frac{20}{144} = \frac{5}{36}\)

Question 13

Two dice are rolled simultaneously and counts are added:

(i) complete the table given below:

Event: ‘Sum on 2 dice’	2	3	4	5	6	7	8	9	10	11	12
Probability	\(\frac{1}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)	\(\frac{4}{36}\)	\(\frac{5}{36}\)	\(\frac{6}{36}\)	\(\frac{5}{36}\)	\(\frac{4}{36}\)	\(\frac{3}{36}\)	\(\frac{2}{36}\)	\(\frac{1}{36}\)

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.

Solution:

(i) Table completed above.

(ii) No, I don’t agree. While there are 11 possible sums, they are not equally likely. For example, there’s only one way to get a sum of 2 (1+1) but six ways to get a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). Therefore, the probability of each sum is different.

Question 14

A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Deskhtha wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that she will lose the game.

Solution:

Total possible outcomes when tossing a coin 3 times = 2³ = 8

Favorable outcomes for winning: HHH, TTT → 2 outcomes

Probability of winning = \(\frac{2}{8} = \frac{1}{4}\)

Probability of losing = 1 – Probability of winning = \(1 – \frac{1}{4} = \frac{3}{4}\)

Question 15

A dice is thrown twice. What is the probability that:

(i) 5 will not come up either time? (ii) 5 will come up at least once?

Solution:

Total possible outcomes when throwing a die twice = 6 × 6 = 36

(i) Outcomes where 5 doesn’t appear either time: Each die has 5 options (1,2,3,4,6). Total = 5 × 5 = 25

Probability = \(\frac{25}{36}\)

(ii) Probability that 5 comes at least once = 1 – Probability that 5 doesn’t appear at all = \(1 – \frac{25}{36} = \frac{11}{36}\)

Exercise 13.1 Solutions – Class X Mathematics

Exercise 13.1 Solutions

Probability – Class X Mathematics

State Council of Educational Research and Training, Telangana, Hyderabad

Question 1: Complete the following statements

(i) Probability of an event \( E + \text{Probability of the event ‘not } E’ = \) ______.

Solution: The sum of the probability of an event and its complement is always 1.

\[ P(E) + P(\text{not } E) = 1 \]

(ii) The probability of an event that cannot happen is ______. Such an event is called ______.

Solution: The probability of an impossible event is 0. Such an event is called an impossible event.

\[ P(\text{Impossible event}) = 0 \]

(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.

Solution: The probability of a certain event is 1. Such an event is called a sure event.

\[ P(\text{Certain event}) = 1 \]

(iv) The sum of the probabilities of all the elementary events of an experiment is ______.

Solution: The sum of probabilities of all elementary events of an experiment is 1.

\[ \sum P(E_i) = 1 \]

(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.

Solution: The probability of any event \( E \) satisfies:

\[ 0 \leq P(E) \leq 1 \]

Question 2: Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

Solution: Not equally likely. The probability depends on the car’s condition, fuel, etc.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Solution: Not equally likely. The outcome depends on the player’s skill.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

Solution: Equally likely. There’s a 50% chance of guessing correctly.

\[ P(\text{Right}) = P(\text{Wrong}) = 0.5 \]

(iv) A baby is born. It is a boy or a girl.

Solution: Equally likely (assuming equal probability for biological sexes).

\[ P(\text{Boy}) = P(\text{Girl}) ≈ 0.5 \]

Question 3: If \( P(E) = 0.05 \), what is the probability of ‘not E’?

Solution:

We know that \( P(E) + P(\text{not } E) = 1 \)

\[ P(\text{not } E) = 1 – P(E) = 1 – 0.05 = 0.95 \]

Question 4: A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag.

(i) an orange flavoured candy?

Solution: Since the bag contains only lemon flavored candies:

\[ P(\text{Orange}) = 0 \] (Impossible event)

(ii) a lemon flavoured candy?

Solution: Since all candies are lemon flavored:

\[ P(\text{Lemon}) = 1 \] (Certain event)

Diagram: A bag containing multiple lemon candies (all yellow) with one being drawn out.

[Illustration of a bag with yellow candies and one hand taking a candy out]

Question 5: Rahim removes all the hearts from the cards. What is the probability of:

i. Getting an ace from the remaining pack.

Solution: Original deck has 52 cards. After removing 13 hearts, 39 cards remain.

Number of aces in remaining cards: 3 (since Ace of Hearts was removed)

\[ P(\text{Ace}) = \frac{3}{39} = \frac{1}{13} \]

ii. Getting a diamonds.

Solution: All 13 diamonds remain in the deck of 39 cards.

\[ P(\text{Diamond}) = \frac{13}{39} = \frac{1}{3} \]

iii. Getting a card that is not a heart.

Solution: Since all hearts have been removed, all remaining 39 cards are not hearts.

\[ P(\text{Not heart}) = \frac{39}{39} = 1 \]

iv. Getting the Ace of hearts.

Solution: Since all hearts have been removed:

\[ P(\text{Ace of Hearts}) = 0 \] (Impossible event)

Diagram: A standard deck of cards with all heart cards (red) removed, showing only clubs, diamonds, and spades remaining.

[Illustration of a partial deck with hearts missing]

Question 6: In a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

Let \( P(\text{Same}) \) be the probability that two students share a birthday.

Given \( P(\text{Not same}) = 0.992 \)

\[ P(\text{Same}) = 1 – P(\text{Not same}) = 1 – 0.992 = 0.008 \]

Question 7: A die is rolled once. Find the probability of getting:

(i) a prime number

Solution: Prime numbers on a die: 2, 3, 5 (3 outcomes)

Total possible outcomes: 6

\[ P(\text{Prime}) = \frac{3}{6} = \frac{1}{2} \]

(ii) a number lying between 2 and 6

Solution: Numbers between 2 and 6: 3, 4, 5 (3 outcomes)

\[ P(\text{Between 2 and 6}) = \frac{3}{6} = \frac{1}{2} \]

(iii) an odd number

Solution: Odd numbers on a die: 1, 3, 5 (3 outcomes)

\[ P(\text{Odd}) = \frac{3}{6} = \frac{1}{2} \]

Diagram: A standard six-faced die showing numbers 1 through 6 with prime numbers (2,3,5) highlighted.

[Illustration of a die with some numbers highlighted]

Question 8: What is the probability of selecting a red king from a deck of cards?

Solution:

Total cards in deck: 52

Red kings: King of Hearts and King of Diamonds (2 cards)

\[ P(\text{Red King}) = \frac{2}{52} = \frac{1}{26} \]

Diagram: A deck of cards showing the two red kings (King of Hearts and King of Diamonds).

[Illustration of two red king cards]

Question 9: Make 5 more problems of this kind using dice, cards or birthdays

Sample Problems:

What is the probability of rolling a number greater than 4 on a standard die?
If two cards are drawn from a standard deck without replacement, what is the probability both are spades?
In a group of 30 people, what is the probability that at least two people share the same birthday?
What is the probability of drawing a face card (Jack, Queen, King) from a standard deck?
If you roll two dice, what is the probability that the sum is 7?

These problems can be discussed with friends and teachers to understand probability concepts better.