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A club has exactly 3 men and 7 women as members. If two memb

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A club has exactly 3 men and 7 women as members. If two memb [#permalink] New post 22 Jul 2018, 08:12
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A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)
[Reveal] Spoiler: OA

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Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink] New post 23 Jul 2018, 07:49
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we can solve this problem by a direct method which is more time consuming.
i.e finding the no of ways that at least one women is selected.
However an easier approach is to find the no of ways in which only men are selected for both the position and reducing this probabilty by 1.

No. of ways in which only men can be selected for both the position is

\(3 * 2 = 6\) since there are \(3\) men and same person cannot hold both the position.
total no of possible outcome would be \(10 * 9 = 90\) This is because men + women = 10

therefore probability = \(\frac{6}{90} = \frac{2}{30} = \frac{1}{15}\)
Hence probability that at least a women is selected = \(\frac{15}{15} - \frac{1}{15} = \frac{14}{15}\)
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Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink] New post 18 Nov 2018, 19:05
amorphous wrote:
we can solve this problem by a direct method which is more time consuming.
i.e finding the no of ways that at least one women is selected.


For understanding, could please explain that too.
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Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink] New post 18 Nov 2018, 19:20
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Carcass wrote:
A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)



OK... the two ways are

1) Cases when none are there..
Total ways = (7+3)(7+3-1)=10*9=90
ways when both are men, => 3*2=6
thus ways when atleast one is woman => 90-6=84
Thus probability that atleast one is woman = \(\frac{84}{90}=\frac{14}{15}\)

2) find each case when women are present
a) only one is present
choose one female = 7C1
choose one male = 3C1
total ways = 7C1*3C1=7*3=21
but the female can be at either of the two position => 21*2=42
b) both are women
first one can be any of 7 and next any of remaining 6, so 7*6=42

Total ways women are present = 42+42=84
probability = \(\frac{84}{90}=\frac{14}{15}\)


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Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink] New post 19 Nov 2018, 00:12
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AE wrote:
amorphous wrote:
we can solve this problem by a direct method which is more time consuming.
i.e finding the no of ways that at least one women is selected.


For understanding, could please explain that too.


At least 1 woman is selected if women and women is selected or women and men is selected or men and women is selected.

total number of possible outcomes is \(10 * 9 = 90\)

w&w = \(7*6 = 42\)
w&m = \(7*3 = 21\)
m&w = \(3*7 = 21\)
therefore the probability that at least 1 woman is selected is \(\frac{84}{90}\)which reduces to \(\frac{14}{15}\)
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Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink] New post 12 Dec 2018, 10:54
We can solve this using combination.
1 - (All Men Probability).

All men probability = 3C2 / 10C2 = 3/45 = 1/15

At least 1 Woman probability = 1-(1/15) = 14/15
Re: A club has exactly 3 men and 7 women as members. If two memb   [#permalink] 12 Dec 2018, 10:54
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