NEOCODE

Probability MCQs

1️⃣ Problems Based on Coins, Dice, and Cards

1. A box contains 4 red, 5 black, and 6 green balls. 3 balls are drawn at random. What is the probability that all the balls are of the same colour?

Correct Answer: a) 33/455

Explanation:
Total number of ways to select 3 balls from 15 balls = 15C3 = 455
Ways to select 3 red balls = 4C3 = 4
Ways to select 3 black balls = 5C3 = 10
Ways to select 3 green balls = 6C3 = 20
Total favorable outcomes = 4 + 10 + 20 = 34
Probability = 34/455
Note: The answer should be 34/455, but option (a) has 33/455 which seems to be a typo in the question.

2. A 4-digit number is formed by the digits 0, 1, 2, 5, and 8 without repetition. Find the probability that the number is divisible by 5.

Correct Answer: c) 3/5

Explanation:
For a 4-digit number, the first digit cannot be 0.
Total digits to choose from = 5 (0, 1, 2, 5, 8)
First position: can use 4 digits (excluding 0) = 4 options
Remaining positions: we select 3 digits from the remaining 4 = 4P3 = 24 ways
Total number of possible 4-digit numbers = 4 × 24 = 96
For a number to be divisible by 5, the last digit must be 0 or 5.
Case 1: Last digit is 0 - First position: 3 options (1, 2, 8) - Middle positions: 3P2 = 6 ways - Total for this case = 3 × 6 = 18 numbers Case 2: Last digit is 5 - First position: 3 options (1, 2, 8) - Middle positions: 3P2 = 6 ways - Total for this case = 3 × 6 = 18 numbers Total favorable outcomes = 18 + 18 = 36
Probability = 36/96 = 3/8
Note: The correct answer should be 3/8, but the given options suggest the answer is 3/5. This could be an error in the question.

3. A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

Correct Answer: a) 175/256

Explanation:
Probability of hitting the target in a single shot = 1/4 = 0.25
Probability of missing the target in a single shot = 3/4 = 0.75
Probability of hitting the target at least once in 4 shots = 1 - Probability of missing all 4 shots
= 1 - (3/4)^4
= 1 - 81/256
= 175/256
Short Trick: This is a binomial probability problem where we need at least 1 success in 4 trials. Calculating the complementary event (missing all 4 shots) makes the solution much simpler.

4. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both hearts. Find the probability of the lost card being a heart.

Correct Answer: c) 11/50

Explanation:
This is a conditional probability problem using Bayes' theorem.
Let H = event that the lost card is a heart
Let B = event that both drawn cards are hearts
We need to find P(H|B) = P(B|H) × P(H) / P(B)
If the lost card is a heart (H):
- Hearts remaining = 12 - Total cards remaining = 51 - P(B|H) = C(12,2)/C(51,2) = 66/1275
If the lost card is not a heart (not H):
- Hearts remaining = 13 - Total cards remaining = 51 - P(B|not H) = C(13,2)/C(51,2) = 78/1275
P(H) = 13/52 = 1/4
P(not H) = 39/52 = 3/4
P(B) = P(B|H) × P(H) + P(B|not H) × P(not H)
= (66/1275) × (1/4) + (78/1275) × (3/4)
= 66/5100 + 234/5100
= 300/5100 = 50/850
P(H|B) = P(B|H) × P(H) / P(B)
= (66/1275) × (1/4) ÷ (50/850)
= (66/1275) × (1/4) × (850/50)
= 66/1275 × 1/4 × 17
= 11/50
Short Trick: Using Bayes' theorem for conditional probability is the most efficient approach for this problem.

5. A bag contains 6 red balls and 8 green balls. 2 balls are drawn at random one by one. Find the probability that both the balls are green.

Correct Answer: a) 16/49

Explanation:
Total balls = 6 + 8 = 14
Drawing balls one by one (without replacement):
P(first ball is green) = 8/14
P(second ball is green | first ball was green) = 7/13
P(both balls are green) = 8/14 × 7/13 = 56/182 = 28/91
Note: The correct answer should be 28/91, not 16/49. There seems to be an error in the options provided.

2️⃣ Conditional Probability

6. 10 persons are seated around a round table. What is the probability that 4 particular persons are always seated together?

Correct Answer: a) 1/21

Explanation:
For a round table, total arrangements = (10-1)! = 9! (because only relative positions matter)
When 4 particular persons sit together, we can treat them as a single unit.
So we have 7 units (1 group of 4 people + 6 individual people)
Number of ways to arrange 7 units in a circle = 6!
The 4 people in the group can be arranged among themselves in 4! ways
Total favorable arrangements = 6! × 4!
Probability = (6! × 4!) / 9!
= (720 × 24) / 362880
= 17280 / 362880
= 1/21
Short Trick: For circular arrangement problems with a group that must stay together, treat the group as one unit and use (n-1)! for total arrangements.

7. A man draws a card from a pack of 52 cards, loses it, and then finds that 2 remaining cards drawn from the deck are both hearts. What is the probability that the lost card was also a heart?

Correct Answer: c) 11/50

Explanation:
This is the same as question 4. Using Bayes' theorem:
Let H = event that the lost card is a heart
Let B = event that both drawn cards are hearts
P(H|B) = P(B|H) × P(H) / P(B)
If the lost card is a heart (H):
- Hearts remaining = 12 - Total cards remaining = 51 - P(B|H) = C(12,2)/C(51,2) = 66/1275
If the lost card is not a heart (not H):
- Hearts remaining = 13 - Total cards remaining = 51 - P(B|not H) = C(13,2)/C(51,2) = 78/1275
P(H) = 13/52 = 1/4
P(not H) = 39/52 = 3/4
P(B) = P(B|H) × P(H) + P(B|not H) × P(not H)
= (66/1275) × (1/4) + (78/1275) × (3/4)
P(H|B) = P(B|H) × P(H) / P(B)
= 11/50
Short Trick: This is identical to question 4, with the same solution and answer.

8. Four boys and three girls stand in queue for an interview. The probability that they stand in alternate positions is:

Correct Answer: c) 1/35

Explanation:
Total number of ways to arrange 7 people = 7! = 5040
For alternating positions, we have two cases:
Case 1: Boys are in positions 1, 3, 5, 7 and girls in positions 2, 4, 6
- Ways to arrange 4 boys among positions 1, 3, 5, 7 = 4! = 24 - Ways to arrange 3 girls among positions 2, 4, 6 = 3! = 6 - Total for Case 1 = 24 × 6 = 144
Case 2: Girls are in positions 1, 3, 5, 7 and boys in positions 2, 4, 6
- This is impossible because we only have 3 girls and 4 positions to fill
Total favorable outcomes = 144
Probability = 144/5040 = 1/35
Short Trick: When we need to place different groups in alternating positions, consider each possible starting group and check if it's valid.

9. A speak truth in 60% cases and B in 80% cases. In what percent of cases are they likely to contradict each other narrating the same incident?

Correct Answer: c) 11/25

Explanation:
A speaks truth: 60% (0.6), A lies: 40% (0.4)
B speaks truth: 80% (0.8), B lies: 20% (0.2)
They will contradict each other in two cases:
Case 1: A speaks truth (0.6) and B lies (0.2)
Probability = 0.6 × 0.2 = 0.12
Case 2: A lies (0.4) and B speaks truth (0.8)
Probability = 0.4 × 0.8 = 0.32
Total probability of contradiction = 0.12 + 0.32 = 0.44 = 11/25
Short Trick: For contradiction, one must tell truth and the other must lie.

3️⃣ Miscellaneous Probability

10. An apartment has 8 floors. An elevator starts with 4 passengers and stops at 8 floors of the apartment. What is the probability that all passengers travel to different floors?

Correct Answer: c) 105/256

Explanation:
Total number of ways 4 passengers can select floors = 8^4 = 4096
(Each passenger has 8 choices independently)
Ways to select different floors = P(8,4) = 8!/(8-4)! = 8!/4! = 1680
Probability = 1680/4096 = 105/256
Short Trick: This is basically asking for the probability of selecting 4 different floors from 8 floors when each passenger makes an independent choice.

11. A man has 9 pairs of dark blue socks and 9 pairs of black socks. He keeps them all in the same bag. If he picks out three socks at random, then what is the probability that he will get a matching pair?

Correct Answer: b) 2×9C2×9C1 / 18C3

Explanation:
Total number of socks = 9×2 + 9×2 = 36 socks (18 pairs)
Total ways to pick 3 socks = C(36,3)
For a matching pair, we need at least 2 socks of the same pair.
For dark blue socks: Select 1 pair (2 socks) from 9 pairs = C(9,1), and 1 sock from remaining 18 black socks = C(18,1)
For black socks: Select 1 pair (2 socks) from 9 pairs = C(9,1), and 1 sock from remaining 18 dark blue socks = C(18,1)
Total favorable outcomes = 2 × C(9,1) × C(18,1)
Probability = (2 × C(9,1) × C(18,1)) / C(36,3)
= (2 × 9 × 18) / (36 × 35 × 34 / 6)
Note: The calculation in option B is not exactly correct. The correct expression should account for the fact that we're selecting from individual socks, not pairs.

12. A box contains 5 red, 7 blue, and 8 green balls. What is the probability that a randomly chosen ball is either red or green?

Correct Answer: a) 13/20

Explanation:
Total number of balls = 5 + 7 + 8 = 20
Number of red balls = 5
Number of green balls = 8
Number of red or green balls = 5 + 8 = 13
Probability = 13/20
Short Trick: For "either A or B" probability (with mutually exclusive events), simply add the individual probabilities.

13. There are four hotels in a town. If 3 men check into the hotels in a day, then what is the probability that each checks into a different hotel?

Correct Answer: b) 1/8

Explanation:
Total number of ways 3 men can check into 4 hotels = 4^3 = 64
(Each man has 4 choices independently)
Ways to check into different hotels = P(4,3) = 4!/(4-3)! = 24
Probability = 24/64 = 3/8
Note: The correct answer should be 3/8 (option C), not 1/8 (option B).