Practice Questions: Probability

Practice Questions: Probability

Q1. A die is thrown then find the probability of getting a perfect square.
(a) 1/3
(b) 1/2
(c) 2/3
(d) 0
Ans: (a)
Sol:
Total possible outcomes on a single fair die = 6.
Perfect squares among {1,2,3,4,5,6} are {1,4}.
Number of favourable outcomes = 2.
Required probability = 2 ÷ 6 = 1 ÷ 3.
Explanation: Two faces are perfect squares out of six equally likely faces, so probability = 1/3.

Q2. If two coins are tossed then find the probability of the event of getting one head .
(a) 1/4
(b) 1/3
(c) 1/2
(d) 3/4
Ans: (c)
Sol:
Sample space for two fair coins = {HH, HT, TH, TT}.
Favourable outcomes for exactly one head = {HT, TH}.
Number of favourable outcomes = 2; total outcomes = 4.
Required probability = 2 ÷ 4 = 1 ÷ 2.
Explanation: Two of the four equally likely outcomes have exactly one head, so probability = 1/2.

Q3. Find the probability of getting 4 when a die is thrown once.
Sol: 
Step 1: Find the probability of getting 4.
Possible outcomes on rolling the die are 1, 2, 3, 4, 5 and 6.
Total number of outcomes = 6.
Number of favourable outcomes = 1 (only the face 4).

Hence, the required probability is 1/6.

Explanation: A fair die has six equally likely faces and only one of them shows 4, so the probability is 1 out of 6.

Q4. In tossing a coin, the chance of throwing head and tail alternatively in 3 successive trials is
(a) 1/4
(b) 1/6
(c) 1/5
(d) 1/48
Ans: (a)
Sol:
Total sequences for three tosses = 2 × 2 × 2 = 8, namely {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
Favourable sequences where heads and tails alternate are HTH and THT → 2 outcomes.
Required probability = 2 ÷ 8 = 1 ÷ 4.
Explanation: Only two of the eight equally likely sequences alternate, so probability = 1/4.

Q5. If three coins are tossed then find the probability of the events of getting exactly one tail.
(a) 1/4
(b) 3/8
(c) 1/2
(d) 5/8
Ans: (b)
Sol:
Total sequences for three fair coins = 8: (H,H,H), (H,H,T), (H,T,H), (T,H,H), (H,T,T), (T,H,T), (T,T,H), (T,T,T).
Favourable outcomes with exactly one tail are (H,H,T), (H,T,H), (T,H,H) → 3 outcomes.
Required probability = 3 ÷ 8.
Explanation: Three of the eight equally likely outcomes have exactly one tail, so probability = 3/8.

Q6. A die is thrown then find the probability of getting an odd number.
(a) 1/3
(b) 2/3
(c) 1/2
(d) None of these
Ans: (c)
Sol:
Sample space S = {1, 2, 3, 4, 5, 6}.
Odd numbers in S are {1, 3, 5}.
Number of favourable outcomes = 3; total outcomes = 6.
Required probability = 3 ÷ 6 = 1 ÷ 2.
Explanation: Three of the six faces are odd, so probability = 1/2.

Q7. A card is drawn at random from well shuffled pack of 52 cards. Find the probability that the card drawn is a spade:
(a) 1/4
(b) 1/3
(c) 1/2
(d) 3/4
Ans: (a)
Sol:
Total number of cards in a standard pack = 52.
Number of spades = 13 (one suit contains 13 cards).
Required probability = 13 ÷ 52 = 1 ÷ 4.
Explanation: One-fourth of the cards are spades, so probability = 1/4.

Q8. Two coins are tossed. Find the probability that a head does not appear:
(a) 1/4
(b) 1/3
(c) 1/2
(d) 2/3
Ans: (a)
Sol:
Sample space for two tosses = {HH, HT, TH, TT}.
Favourable outcome where no head appears = {TT}.
Number of favourable outcomes = 1; total outcomes = 4.
Required probability = 1 ÷ 4.
Explanation: Only TT has no head, so probability = 1/4.

Q9. A box contains 20 cards marked with the numbers 1 to 20. One card is drawn from this box at random. What is the probability that number on the card is a multiple of 5?
(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5
Ans: (a)
Sol: 
Total number of cards = 20.
Multiples of 5 between 1 and 20 are {5, 10, 15, 20} → 4 favourable outcomes.
Required probability = 4 ÷ 20 = 1 ÷ 5.
Explanation: Four numbers out of twenty are multiples of 5, so probability = 1/5.

Q10. If two coins are tossed then find the probability of the events that at the most one tail turns up
(a) 1/4
(b) 1/3
(c) 1/2
(d) 3/4
Ans: (d)
Sol:
Sample space for two coins = {HH, HT, TH, TT}.
"At most one tail" means 0 or 1 tail, so favourable outcomes are {HH, HT, TH} → 3 outcomes.
Required probability = 3 ÷ 4 = 3/4.
Explanation: Three of the four equally likely outcomes have at most one tail, so probability = 3/4.

Q11. If three coins are tossed then find the probability of the event of getting no tail.
(a) 1/8
(b) 1/4
(c) 3/8
(d) 1/2
Ans: (a)
Sol:
Total possible outcomes for three coins = 8.
"No tail" means all three are heads: (H, H, H) → 1 favourable outcome.
Required probability = 1 ÷ 8.
Explanation: Only one of the eight equally likely outcomes has no tail, so probability = 1/8.

Q12. A box contains 3 red, 3 white and 3 green balls. A ball is selected at random. Find the probability that the ball picked up is neither a white nor a red ball:
(a) 1/4
(b) 1/3
(c) 1/2
(d) 3/4
Ans: (b)
Sol:
Total number of balls = 3 + 3 + 3 = 9.
Balls that are neither white nor red are the green balls = 3.
Required probability = 3 ÷ 9 = 1 ÷ 3.
Explanation: One third of the balls are green, so probability = 1/3.

Q13. Two die are thrown find the probability of getting the sum of the numbers on their upper faces to be at most 3.
(a) 1/4
(b) 1/6
(c) 1/12
(d) 1/36
Ans: (c)
Sol:
Total ordered outcomes when two dice are thrown = 6 × 6 = 36.
Ordered pairs (a, b) with sum at most 3 are (1,1), (1,2), (2,1) → 3 outcomes.
Required probability = 3 ÷ 36 = 1 ÷ 12.
Explanation: Three of the 36 equally likely ordered pairs give sum ≤ 3, so probability = 1/12.

Q14. Two die are thrown. Find the probability that the number on the upper face of the first dice is less than the number on the upper face of the second dice.
(a) 1/2
(b) 7/12
(c) 1/3
(d) 5/12
Ans: (d)
Sol:
Total ordered pairs = 6 × 6 = 36.
Count ordered pairs where first < />
If first = 1 then second can be 2-6 → 5 possibilities.
If first = 2 then second can be 3-6 → 4 possibilities.
If first = 3 then second can be 4-6 → 3 possibilities.
If first = 4 then second can be 5-6 → 2 possibilities.
If first = 5 then second can be 6 → 1 possibility.
If first = 6 then no possibility → 0.
Total favourable = 5 + 4 + 3 + 2 + 1 = 15.
Required probability = 15 ÷ 36 = 5 ÷ 12.
Explanation: Out of 36 ordered outcomes, 15 have the first die less than the second, so probability = 5/12.

Q15. If two coins are tossed then find the probability of the event that no head turns up.
(a) 1/4
(b) 1/3
(c) 1/2
(d) 3/4
Ans: (a)
Sol:
Sample space = {HH, HT, TH, TT}.
"No head" means both are tails: TT → 1 favourable outcome.
Required probability = 1 ÷ 4.
Explanation: Only one of the four equally likely outcomes has no head, so probability = 1/4.