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Problem:

Alex flip a fair coin three times. what is the probability that she gets two heads given that the first is a head?


My solution is based on the argument that from the given information the sample space can be limited to four elements {$HHH,HHT,HTH,HTT$} Of these {$HHH,HHT,HTH$} satisfies the required event.Thus the probability is equal to $\frac{3}{4} = 0.75$.

But the given answer is $0.5$,where exactly is the problem in my approach ?

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Maybe I just don't see the problem, but flipping a coin once, twice or multiple times won't change the fact that the probabilities for heads and tails are $0.5$. Also, the coin doesn't know that the first flip was heads. So you simply calculate the probability that you get one head with two flips which is $0.5$. –  Huy Dec 12 '10 at 12:04

2 Answers 2

up vote 2 down vote accepted

Given that the first flip is a head, the problem is equivalent to asking, what is the probability of getting exactly $1$ head in $2$ flips. There are two possibilities, that the first flip is a head, and the second is tails, and vice versa. This gives the probability the given answer says.


To address your question more directly, you should use conditional probability. Let $A$ be the event that Alex flips exactly $2$ heads, and $B$ be the event that the first flip is heads.

Based on the sample space you wrote out, note $P(A\cap B)=2/8$, as the two possibilities are $\{HHT,HTH\}$. Also, $P(B)=1/2$, as you listed the four possibilities.

Now use the formula

$$P(A|B)=\frac{P(A\cap B)}{P(B)}.$$

I think your main error is that $\{HHH\}$ does not satisfy the condition that the first flip is a head, and that there are exactly two heads, and thus should not be included when calculating the probability $P(A\cap B)$.

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What I am looking for is the problem with my approach. –  Quixotic Dec 12 '10 at 12:06
1  
You counted the possibility of having three heads. But I think the question asks for two heads in total. –  Raskolnikov Dec 12 '10 at 12:11
    
Yes I have already tried and solved with that formula,but don't you think the problem statement should mention the word exactly for the case it want us to neglect the condition with three heads? –  Quixotic Dec 12 '10 at 12:15
    
Also I would I like to add the conditional probability approach is a bit lengthy if you can limit the sample space ;-) –  Quixotic Dec 12 '10 at 12:19
1  
@Debanjan, you are right that the language is not entirely clear. I think if the author of the problem wanted you to include that case, the question would have specifically asked for the probability of Alex getting at least $2$ heads, as opposed to just saying Alex gets $2$ heads, which is more often than not assumed to mean exactly $2$ heads. –  yunone Dec 12 '10 at 12:19

The problem here is that if the answer is half, then what the question means is exactly two heads, so three heads are not allowed. In your solution you included the case with three heads.

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But the word exactly is missing! –  Quixotic Dec 12 '10 at 12:16
    
+1,The setter is way out of my reach...anyways thanks for your time :) –  Quixotic Dec 12 '10 at 12:23
    
That's very true! –  Quixotic Dec 12 '10 at 12:30

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