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by Aishwarya1204 » Fri Oct 19, 2012 12:16 pm
If a fair coin is tossed 3 times, what is the probability that it will turn up heads exactly twice?

answer : [spoiler] 3/8[/spoiler]

Is there any way of doing this without having to list out the possibilities like
HHH HHT HTT THT etc.
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by GMATGuruNY » Fri Oct 19, 2012 12:45 pm
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by mariofelixpasku » Fri Oct 19, 2012 1:04 pm
imagine it like this Yes Yes No (YYN)

then we would have 1/2 * 1/2 * 1/2 = 1/8

but we are not done yet cause we have to multiply it with the number of possibilities the 2 Y can be arranged among the 3 spots of YYN.

therefore we have 1/8 * 3P3/2! = 1/8 * 3!/2! = 3/8
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by Brent@GMATPrepNow » Sat Oct 20, 2012 7:26 am
Aishwarya1204 wrote:If a fair coin is tossed 3 times, what is the probability that it will turn up heads exactly twice?

answer : [spoiler] 3/8[/spoiler]

Is there any way of doing this without having to list out the possibilities like
HHH HHT HTT THT etc.
I think that listing is a great way to solve this question. Too many students discount this approach.

Here's a more mathematical approach:
P(exactly 2 heads) = P(THH OR HTH OR HHT)
= P(THH) + P(HTH) + P(HHT)
= (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2)
= (1/8) + (1/8) + (1/8)
= [spoiler]3/8[/spoiler]

Cheers,
Brent
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