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I have a very simple conditional probability question that I can't reason through.

A coin is tossed three times what is the probability exactly two heads occur given that:

a) The first outcome was a head

The answer is obviously 1/2, since the remaining outcomes are {TT,TH,HT,HH} so there's a 1/2 chance of exactly two heads. But I'm struggling to do it using the conditional probability formula:

P(2H n 1 heads) /Prob (1 head)

I get (3/8)/(1/2)=3/4 whic isn't the right answer. Can someone use the conditional probability formula to solve this simple question?

Also how would the formula be applied if the first toss was a heads?

Pedro
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William Smith
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4 Answers4

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Let $\{H^2T\}_{1..3}$ be the event of getting two heads and a tail, in any order, in trials 1,2,3. And so forth.

ie: $\{H^2T\}_{1..3} \equiv \{(H,H,T), (H,T,H), (T,H,H)\}$

So, $\mathrm{P}(\{H^2T\}_{1..3}\mid \{H\}_1) $ $ = \dfrac{\mathrm{P}(\{H\}_1\cap\{H^2T\}_{1..3})}{\mathrm{P}(\{H\}_1)} \\ = \dfrac{\mathrm{P}(\{H\}_1\cap\{HT\}_{2,3})}{\mathrm{P}(\{H\}_1)}\\ = \dfrac{\mathrm{P}(\{H\}_1)\mathrm{P}(\{HT\}_{2,3})}{\mathrm{P}(\{H\}_1)}\\ = \mathrm{P}(\{HT\}_{2,3}) \\ = {2\choose 1}\mathrm{P}(H)\mathrm{P}(T)\\ = \frac 12$

Graham Kemp
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The Pr(2H in all rounds|1st round I got a H)

= Pr(2H in all rounds $\cap$ H in 1st round)/Pr(H in 1st round)

Now Pr(2H in all rounds $\cap$ H in 1st round) means (HHT/HTH). So, this Prob. is 2/8 = 1/4

Thus Pr(2H in all rounds|1st round I got a H) = (1/4)/(1/2) = 1/2

Sayan Bandyapadhyay
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P(2H n 1 head)=1/8 P(2H)=2/8 sample space={HHH,HTT,THT,TTH,HHT,HTH,TTH,TTT} consider the subset {HHH} for P(2H n 1 head), and {HTH,HHT} for P(2H)

abstract
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The event "get exactly two heads in three coin flips and the first coin flip lands heads" has two outcomes ($\{HTH,HHT\}$) out of the possible 8 outcomes for three coin flips.

The event "the first coin flip lands heads" has four outcomes ($\{HTT,HTH,HHT,HHH\}$) out of the possible 8 outcomes for three coin flips.

The probability that you get exactly two heads in three coin flips ($A$) given that the first coin flip lands heads ($B$): $P[A|B]=P[A\cap B]\div P[B]=\frac28\div \frac48=\frac12$

JRN
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