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?

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?

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?

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:

(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 = \(1 – \frac{25}{36} = \frac{11}{36}\)