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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink]
Carcass i request an explanation
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink]
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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]
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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)
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Re: In a party there were 80 males and 45 females. One pair of a male and [#permalink]
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