Question 1: Complete the following statements:

(i) Probability of an event E + Probability of the event 'not E' = _____ .
Solution: 1

(ii) The probability of an event that cannot happen is _____ . Such an event is called _____ .
Solution: 0, Impossible event

(iii) The probability of an event that is certain to happen is _____ . Such an event is called _____ .
Solution: 1, Sure or Certain event

(iv) The sum of the probabilities of all the elementary events of an experiment is _____ .
Solution: 1

(v) The probability of an event is greater than or equal to _____ and less than or equal to _____ .
Solution: 0, 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 car starting depends on various factors like battery, fuel, engine condition, etc.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Solution: Not equally likely. It depends on the player's skill and training.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Solution: Equally likely. There are only two outcomes and both have an equal chance of being selected.

(iv) A baby is born. It is a boy or a girl.
Solution: Equally likely. Naturally, both outcomes have equal probability.

Question 3: Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution: When a coin is tossed, the outcomes 'Head' and 'Tail' are equally likely. Since the result is completely unpredictable and unbiased, tossing a coin is considered a fair decision-making method.

Question 4: Which of the following cannot be the probability of an event?

(A) 2/3    (B) -1.5    (C) 15%    (D) 0.7

Solution: (B) -1.5
Reason: Probability of an event can never be negative, i.e., 0 ≤ P(E) ≤ 1.

Question 5: If P(E) = 0.05, what is the probability of 'not E'?

Solution:
We know that P(E) + P(not E) = 1
P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.

Question 6: A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:

(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Solution:
(i) Orange candy: The bag contains NO orange candies. It is an impossible event. Therefore, P(orange candy) = 0.
(ii) Lemon candy: The bag contains ONLY lemon candies. It is a sure event. Therefore, P(lemon candy) = 1.

Question 7: It is given that 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 E be the event of two students having the same birthday.
Given: P(not E) = 0.992
P(E) = 1 - P(not E) = 1 - 0.992 = 0.008.

Question 8: A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Solution:
Total number of balls = 3 + 5 = 8.
(i) P(red ball) = Number of red balls / Total balls = 3/8.
(ii) P(not red) = 1 - P(red) = 1 - 3/8 = 5/8.

Question 9: 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 + 8 + 4 = 17.
(i) P(red) = 5/17.
(ii) P(white) = 8/17.
(iii) Green marbles = 4, so non-green marbles = 17 - 4 = 13.
P(not green) = 13/17.

Question 10: A piggy bank contains hundred 50p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 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 50 p coin? (ii) will not be a Rs 5 coin?

Solution:
Total number of coins = 100 + 50 + 20 + 10 = 180.
(i) Number of 50p coins = 100
P(50p coin) = 100 / 180 = 5/9.
(ii) Number of Rs 5 coins = 10.
Coins that are not Rs 5 = 180 - 10 = 170.
P(not Rs 5 coin) = 170 / 180 = 17/18.

Question 11: 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 + 8 = 13.
Number of male fish = 5.
P(male fish) = 5/13.

Question 12: 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. 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) Favorable outcome for 8 = 1 ⇒ P(8) = 1/8.
(ii) Odd numbers (1, 3, 5, 7) = 4 outcomes ⇒ P(odd) = 4/8 = 1/2.
(iii) Numbers greater than 2 (3, 4, 5, 6, 7, 8) = 6 outcomes ⇒ P(>2) = 6/8 = 3/4.
(iv) Numbers less than 9 (1, 2, 3, 4, 5, 6, 7, 8) = 8 outcomes ⇒ P(<9) = 8/8 = 1.

Question 13: A die is thrown once. Find the probability of getting:

(i) a prime number
(ii) a number lying between 2 and 6
(iii) an odd number

Solution:
Total outcomes = {1, 2, 3, 4, 5, 6} = 6.
(i) Prime numbers = {2, 3, 5} (3 outcomes) ⇒ P(prime) = 3/6 = 1/2.
(ii) Numbers between 2 and 6 = {3, 4, 5} (3 outcomes) ⇒ P(between 2 and 6) = 3/6 = 1/2.
(iii) Odd numbers = {1, 3, 5} (3 outcomes) ⇒ P(odd) = 3/6 = 1/2.

Question 14: 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

Solution:
Total cards = 52.
(i) Red kings = 2 (King of hearts, King of diamonds) ⇒ P(red king) = 2/52 = 1/26.
(ii) Face cards = 12 (4 Jacks, 4 Queens, 4 Kings) ⇒ P(face card) = 12/52 = 3/13.
(iii) Red face cards = 6 ⇒ P(red face card) = 6/52 = 3/26.
(iv) Jack of hearts = 1 ⇒ P(jack of hearts) = 1/52.
(v) Spades = 13 ⇒ P(spade) = 13/52 = 1/4.
(vi) Queen of diamonds = 1 ⇒ P(queen of diamonds) = 1/52.

Question 15: 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?

Solution:
Total cards initially = 5.
(i) Number of queens = 1 ⇒ P(queen) = 1/5.
(ii) Queen is removed, remaining cards = 4.
(a) Number of aces = 1 ⇒ P(ace) = 1/4.
(b) Number of queens left = 0 ⇒ P(queen) = 0/4 = 0.

Question 16: 12 defective pens are accidentally mixed with 132 good ones. One pen is taken out at random. Determine the probability that the pen taken out is a good one.

Solution:
Good pens = 132, Defective pens = 12.
Total pens = 132 + 12 = 144.
P(good pen) = 132 / 144 = 11/12.

Question 17: (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random. What is the probability that this bulb is defective?

Solution:
Total bulbs = 20, Defective = 4.
P(defective) = 4/20 = 1/5.

(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?
Remaining total bulbs = 19.
Remaining non-defective bulbs = 16 - 1 = 15.
P(not defective) = 15/19.

Question 18: A box contains 90 discs numbered from 1 to 90. Find the probability that a disc drawn 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 (Total 81 numbers) ⇒ P(two-digit) = 81/90 = 9/10.
(ii) Perfect squares = {1, 4, 9, 16, 25, 36, 49, 64, 81} (9 numbers) ⇒ P(perfect square) = 9/90 = 1/10.
(iii) Divisible by 5 = {5, 10, 15, ..., 90} (18 numbers) ⇒ P(divisible by 5) = 18/90 = 1/5.

Question 19: A die has faces showing letters: A, B, C, D, E, A. Find the probability of getting (i) A (ii) D.

Solution:
Total faces = 6.
(i) Number of faces with 'A' = 2 ⇒ P(A) = 2/6 = 1/3.
(ii) Number of faces with 'D' = 1 ⇒ P(D) = 1/6.

Question 20: Suppose you drop a die at random on a rectangular region of dimensions 3m × 2m. What is the probability that it will land inside the circle with diameter 1m?

Solution:
Area of rectangular region = 3m × 2m = 6 sq m.
Diameter of circle = 1m ⇒ Radius (r) = 1/2 m.
Area of circle = πr² = π × (1/2)² = π/4 sq m.
P(landing inside circle) = Area of circle / Total Area = (π/4) / 6 = π/24.

Question 21: 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. What is the probability that:

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

Solution:
Total pens = 144.
Defective pens = 20, Good pens = 144 - 20 = 124.
(i) P(Buy pen) = P(Good pen) = 124 / 144 = 31/36.
(ii) P(Not buy pen) = P(Defective pen) = 20 / 144 = 5/36.

Question 22: Two dice are thrown at the same time. Find the probability of getting a sum of 8 on two dice.

Solution:
Total possible outcomes when two dice are thrown = 6 × 6 = 36.
Favorable outcomes for sum = 8 are {(2,6), (3,5), (4,4), (5,3), (6,2)} = 5 outcomes.
P(sum of 8) = 5/36.

Question 23: A game consists of tossing a one rupee coin 3 times. Hanif wins if all tosses give the same result (3 Heads or 3 Tails), and loses otherwise. Calculate the probability that Hanif will lose.

Solution:
Total outcomes = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} = 8 outcomes.
Winning outcomes = {HHH, TTT} = 2 outcomes.
Losing outcomes = 8 - 2 = 6 outcomes.
P(Hanif loses) = 6/8 = 3/4.

Question 24: 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?

Solution:
Total outcomes = 36.
(i) Outcomes where 5 comes up at least once = {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6)} = 11 outcomes.
Outcomes where 5 does NOT come up = 36 - 11 = 25 outcomes.
P(5 will not come up either time) = 25/36.
(ii) P(5 will come up at least once) = 11/36.