Question 9

Mathematics Probability Hard

A box contains 2 blue caps, 4 red caps, 5 greens caps and 1 yellow cap. If four caps are picked at random, the probability that none of the is green is

(A) 7/99
(B) 7/12
(C) 5/99
(D) 5/12
✓ Correct Answer: Option A Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

The total number of caps: \[ 2 + 4 + 5 + 1 = 12 \] The number of ways for selecting four caps: \[ ^{12}C_4 \] The total number of caps other than green: \[ 12 - 5 = 7 \] The number of ways for selecting four caps other than green: \[ ^{7}C_4 \] The probability of selecting four caps and none of them are green \( P(A') \): \[ P(A') = \frac{\text{Number of ways for selecting four caps other than green}}{\text{Number of ways for selecting four caps}} \] \[ P(A') = \frac{^{7}C_4}{^{12}C_4} \] \[ P(A') = \frac{\frac{7!}{(7-4)!4!}}{\frac{12!}{(12-4)!4!}} \] \[ P(A') = \frac{7!}{(7-4)!4!} \times \frac{(12-4)!4!}{12!} \] \[ P(A') = \frac{7}{99} \]

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