Among 15 players, 8 are batsman and 7 are bowlers. Find the probabilit

1.	Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ?
Answer» The chosen consists of players (6 + 5). ∴ Number of ways of selecting 11 players out of 15 players = n(S) = ¹⁵C₁₁ Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers Then, n(A) = ⁸C₆ x ⁷C₅ ∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{^8C_6\times^7C_5}{^{15}C_{11}}\) = \(\frac{\frac{8\times7}{2}\times\frac{7\times6}{2}}{\frac{15\times14\times13\times12}{4\times3\times3\times1}}\)= \(\frac{8\times7}{2}\) x \(\frac{7\times6}{2}\) x \(\frac{4\times3\times2\times1}{15\times14\times13\times12}\) = \(\frac{28}{65}.\)

Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ?

Answer»

The chosen consists of players (6 + 5).

∴ Number of ways of selecting 11 players out of 15 players = n(S) = ¹⁵C₁₁

Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers

Then, n(A) = ⁸C₆ x ⁷C₅

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{^8C_6\times^7C_5}{^{15}C_{11}}\) = \(\frac{\frac{8\times7}{2}\times\frac{7\times6}{2}}{\frac{15\times14\times13\times12}{4\times3\times3\times1}}\)= \(\frac{8\times7}{2}\) x \(\frac{7\times6}{2}\) x \(\frac{4\times3\times2\times1}{15\times14\times13\times12}\) = \(\frac{28}{65}.\)

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