There are 6 yellow, 5 blue, and 11 red balls in a sack (total 22 balls).
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There are 6 yellow, 5 blue, and 11 red balls in a sack (total 22 balls). Three balls are drawn one by one without replacement.
Find the probability such that the 1st ball is red, the 2nd is yellow, and the 3rd is blue.
Show answer & explanation
Answer: A. 0.0357142857143
P(Red then Yellow then Blue) = (11/22) * (6/21) * (5/20) = (1/2) * (2/7) * (1/4) = 1/28 ≈ 0.0357142857143.
Step-by-step Derivation:
Step 1: Total balls = 6 + 5 + 11 = 22.
Step 2: P(1st is Red) = 11/22 = 1/2.
Step 3: P(2nd is Yellow | 1st Red) = 6/21 = 2/7.
Step 4: P(3rd is Blue | 1st Red, 2nd Yellow) = 5/20 = 1/4.
Step 5: Product = (1/2) * (2/7) * (1/4) = 1/28 ≈ 0.0357142857143.