Question

A committee has to be made of 5 members from 6 men and 4 women. The probability that at least one woman is present in committee, is

• A. $\frac { 1 } { 42 }$
• B. $\frac { 41 } { 42 }$
• C. $\frac { 2 } { 63 }$
• D. $\frac { 1 } { 7 }$

Answer 1

Answer: (B)

$\frac { 41 } { 42 }$

Answer 2

Total number of ways

$= { } ^ { 4 } C _ { 1 } \times { } ^ { 6 } C _ { 4 } + { } ^ { 4 } C _ { 2 } \times { } ^ { 6 } C _ { 3 } + { } ^ { 4 } C _ { 3 } \times { } ^ { 6 } C _ { 2 } + { } ^ { 4 } C _ { 4 } \times { } ^ { 6 } C _ { 1 } + { } ^ { 6 } C _ { 5 }$

$= 60 + 120 + 60 + 6 + 6 = 252$

No. of ways in which at least one woman exist are

$= { } ^ { 4 } C _ { 1 } \times { } ^ { 6 } C _ { 4 } + { } ^ { 4 } C _ { 2 } \times { } ^ { 6 } C _ { 3 } + { } ^ { 4 } C _ { 3 } \times { } ^ { 6 } C _ { 2 } + { } ^ { 4 } C _ { 4 } \times { } ^ { 6 } C _ { 1 } = 246$

Hence required probability $= \frac { 246 } { 252 } = \frac { 41 } { 42 }$ .