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In a club, 75% of the members are men, and 50% of them are the only...

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In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 12 Nov 2018, 18:35
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Question Stats:

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In a club, 75% of the members are men, and 50% of them are the only engineers in the club. If total number of members in the club is 40 and two of them are selected at random. Then what is the probability that at least one of them is a male or an engineer?

    A. \(\frac{9}{156}\)
    B. \(\frac{24}{39}\)
    C. \(\frac{3}{4}\)
    D. \(\frac{7}{8}\)
    E. \(\frac{147}{156}\)

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Re: In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 13 Nov 2018, 01:12
1
No of male non E (M) = 15
No of male E (ME)= 15
No of female (F)= 10
No of ways to select 2 people = 40c2 = 780\

No of ways to select at least 1 male or engineer =
1M x 1F = 15c1 x 10c1 = 150
1ME x 1F = 15c1 x 10c1 = 150
1M x 1ME = 15c1 x 15c1 = 225
2M = 15c2 = 105
2ME = 15c2 = 105

735/780 = 147/156

Option E
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Re: In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 14 Nov 2018, 04:15
1
3

Solution


Given:
    • Total members of the club = 40
    • 75% of the total members are men = 0.75 * 40 = 30
    • 50% of the men are engineers = 0.5 * 30 = 15, and these are the only engineers in the club

To find:
    • If two members are selected at random, what is the probability that at east one of them is a male or an engineer

Approach and Working:
    • P(at least one male or engineer) = 1 – P(no male or engineer)
      o P(no male or engineer) = P(selecting two women) = \(\frac{^{10}C_2}{^{40}C_2} = \frac{9}{156}\)

    • Thus, P(at least one male or engineer) = \(1- \frac{9}{156} = \frac{147}{156}\)

Hence the correct answer is Option E.

Answer: E

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Re: In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 14 Nov 2018, 22:06
Hi guys,
Can anyone explain the ostensible logical confusion in the answer strategy.
The previous answer used 1 - P(Neither) Logic.
This would mean that the calculation is for either scenario (one can visualize easily in a Venn Diagram )
In this particular case , as you know the engineer area or circle is completely included within the Men's circle,since all engineers are men.
The entire intersection is part of Engineer circle.
The either circle is the Men's circle (Please correct me here if I am wrong)
So I calculated it as 30C2 / 40C2 and got it as 87/156.

My question: Why is this different from 1- P(Neither case).
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In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 14 Nov 2018, 22:21
kiranteja93 wrote:
Hi guys,
Can anyone explain the ostensible logical confusion in the answer strategy.
The previous answer used 1 - P(Neither) Logic.
This would mean that the calculation is for either scenario (one can visualize easily in a Venn Diagram )
In this particular case , as you know the engineer area or circle is completely included within the Men's circle,since all engineers are men.
The entire intersection is part of Engineer circle.
The either circle is the Men's circle (Please correct me here if I am wrong)
So I calculated it as 30C2 / 40C2 and got it as 87/156.

My question: Why is this different from 1- P(Neither case).


Hi kiranteja93,

You have missed one case, where we can select one from the 30 members, who are male, set and the other member from the remaining 10 members set. Since, we are asked to find out the probability of at least one male or engineer.

The probability of selecting one member from 30 set and other from the remaining 10 set would be \(^{30}C_1 * ^{10}C_1/^{40}C_2 = \frac{60}{156}\)

Thus, answer = \(\frac{87}{156} + \frac{60}{156} = \frac{147}{156}\)

Regards,
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Re: In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 14 Nov 2018, 22:30
EgmatQuantExpert wrote:
In a club, 75% of the members are men, and 50% of them are the only engineers in the club. If total number of members in the club is 40 and two of them are selected at random. Then what is the probability that at least one of them is a male or an engineer?

    A. \(\frac{9}{156}\)
    B. \(\frac{24}{39}\)
    C. \(\frac{3}{4}\)
    D. \(\frac{7}{8}\)
    E. \(\frac{147}{156}\)

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Total Members = 40
Total Males = 75% of 40 = 30
Total Engineers = Total male Engineers = 50% of males = 15
Total Females = 10

probability that at least one of them is a male or an engineer = Probability that at least one of them is a male = MM + MF
= 30C2/40C2 + 30C1*10C1/40C2
= (30*29/2)/(40*39/2) + 30*10/(40*39/2)
= 435/780 + 300/780
= 735/780
= 147/156

Answer E
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Re: In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 15 Nov 2018, 13:08
Hi Payal,
Thanks for the response.The learning here is subtle : Atleast one case is different from either case which only dances around in the two circles without including neither area as part of possibilities...Interesting.Thankyou
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Re: In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 29 Nov 2018, 09:12
why is the probability of selecting two women equal to P(selecting two women) = 10C2 / 40C2= 9/156 and not 1/4*1/4 = 1/16 ?
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Re: In a club, 75% of the members are men, and 50% of them are the only...  [#permalink]

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New post 29 Nov 2018, 12:46
Zynxu wrote:
why is the probability of selecting two women equal to P(selecting two women) = 10C2 / 40C2= 9/156 and not 1/4*1/4 = 1/16 ?



They are not independent events. You can calculate your way but after you have already selected one woman, you then have 9 left out of the remaining 39 people.

1/4 * 9/39 = 9/156
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Re: In a club, 75% of the members are men, and 50% of them are the only... &nbs [#permalink] 29 Nov 2018, 12:46
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