CMAT 2015 Solved Question Paper

Question : 19 of 100

Marks: +1, -0

What is the probability of choosing 3 distinct numbers randomly from (1, 2, 3, ...,100) such that all are divisible by both 2 and 3?

Solution:

(d) Number of number divisible by both 2 and 3 between 1 to 100 =16
∴ Required probability =

^16c_3/{^100c_3}

{16!}/{3!13!}×{3! × 97!}/{100!}

\;\;\;\;\;\;\;\;\;\;\;\;\;\;

{16⋅15⋅14}/{3!} × {3!}/{100⋅99⋅98}

4/1155

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