A committee of 4 persons is to be appointed from 7 men and 3 women. What is the':
probability that the committee contains (i) exactly two women and (ii) at least one woman?
Answers
Answer:
It is given that there are 4 men and 3 women which means the total number of person are 7.
The number of ways of selecting five person is:
n(S)=
7
C
5
=
(7−5)!×5!
7!
=
2!×5!
7!
=
2×1×5!
7×6×5!
=7×3=21
Let A denote the event in which a committee of 5 is selected as to have at least two men, therefore,
n(A)=(
4
C
2
×
3
C
3
)+(
4
C
3
×
3
C
2
)+(
4
C
4
×
3
C
1
)
=(
(4−2)!×2!
4!
×
(3−3)!×3!
3!
)+(
(4−3)!×3!
4!
×
(3−2)!×2!
3!
)+(
(4−4)!×4!
4!
×
(3−1)!×1!
3!
)
=(
2!×2!
4!
×
0!×3!
3!
)+(
1!×3!
4!
×
1!×2!
3!
)+(
0!×4!
4!
×
2!×1!
3!
)
=(
2×1×2!
4×3×2!
×
1×3!
3!
)+(
1×3!
4×3!
×
1×2!
3×2!
)+(
1×4!
4!
×
1×2!
3×2!
)
=(6×1)+(4×3)+(1×3)=6+12+3=21
Thus, probability that the committee will have at least two men is:
P(A)=
n(S)
n(A)
=
21
21
=1
Hence, probability that the committee will have at least two men is 1.
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