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In a party there were 80 males and 45 females. One pair of a male and

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sg20
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In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   01 Dec 2018, 01:36
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In a party there were 80 males and 45 females. One pair of a male and a female was selected at random for dance, what is the probability that a pair selected is a married couple if there were total 30 married couples in the party?

a) 1/30
b) 1/3600
c) 1/1200
d) 1/120
e) none of these
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   01 Mar 2019, 06:14
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Assuming 30 women are married to 30 men: imagine you pick the woman first, then the man. When you pick the woman, you must pick one of the married women to have any hope of getting a married couple. 30/45 = 2/3 of the women are married, so that's the probability our first selection is 'good'. Then once we've picked the married woman, we need to pick the specific man she is married to. There's 1 man she's married to, out of 80 men in total, so a 1/80 probability we pick the 'right' man. Since we need both of these things to happen, we multiply the probabilities:

(2/3) (1/80) = 1/120
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   01 Dec 2018, 11:12
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As there are 30 couples , the no of ways we can select 1 pair of married couple : 30C1= 30
Now the he no of ways we can select 1 man from 80=80C1= 80
& the he no of ways we can select 1 woman from 45=45C1= 45
Hence , the probability = 30/(80*45)= 1/120............... Ans D.
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   01 Dec 2018, 01:48
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This question doesn't work now on account of gay marriage. It requires 30/45 of the females to be married to 30/80 of the males.

2/3·1/80=1/120
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In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   Updated on: 01 Mar 2019, 09:42
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Male =80 & female= 45

married couple = 30
so
married males; 30 and females 30

we can say
P of married males ; 30/80
and them with females ; 1/45
30/80 * 1/45
IMO 1/120 ; D

Originally posted by Archit3110 on 01 Mar 2019, 05:51.
Last edited by Archit3110 on 01 Mar 2019, 09:42, edited 1 time in total.
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   23 Mar 2020, 10:49
sg20 wrote:
In a party there were 80 males and 45 females. One pair of a male and a female was selected at random for dance, what is the probability that a pair selected is a married couple if there were total 30 married couples in the party?

a) 1/30
b) 1/3600
c) 1/1200
d) 1/120
e) none of these


Doesn't look like a 700 question tbh.
You need to pick one couple from the 30 married, which can be done in 30C1 ways. Let's get to the denominator, you need to form a *straight* couple, that means you need to pick 1 from the males and 1 from the females to qualify as a couple. This can be done in 80C1*45C1 ways.

So IMO the probability will be 30/80*45 = 1/120.

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In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   13 Jul 2020, 04:44
Bunuel VeritasKarishma

When they ask for the probability of selecting a couple, why can't we do 60C1 x 2 in the numerator? (60 ways of selecting the first person and then since the person can be either male or female - 2 ways of selecting the first person). If not in this question, can you also please elaborate when can we use this technique.
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   13 Jul 2020, 04:49
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crackGMAT760 wrote:
Bunuel VeritasKarishma

When they ask for the probability of selecting a couple, why can't we do 60C1 x 2 in the numerator?


crackGMAT760

60C1*2 = 120

While there are only 30 ways of choosing a couple out of 30 couples.

therefore 60C1*2 doesn't have merit.

To. make it slightly correct

60C1 = Number of ways of choosing one person out of 30 couples (60 individuals)

But now, the second person can be chosen in ONLY 1 WAY because chosen person has only 1 partner who makes couple with chosen one.

so you don't need to multiply it by 2

in that case the method will be 60C1*1/(80*45*2)

Denominator also must include arrangement if Numerator includes that (60C1 includes arrangement)

I hope this help! :)
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   13 Jul 2020, 05:01
sg20 wrote:
In a party there were 80 males and 45 females. One pair of a male and a female was selected at random for dance, what is the probability that a pair selected is a married couple if there were total 30 married couples in the party?

a) 1/30
b) 1/3600
c) 1/1200
d) 1/120
e) none of these


Chances of selecting a married couple = 30
Chances of selecting a male = 80
Chances of selecting a female = 45

Hence, probability = 30/(80*45) = 1/120

OA: D
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink] New post   05 Feb 2024, 08:56
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