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Suppose $20$ tosses of a coin yield $8$ heads, $12$ tails.

Let $X$ be the # of sequences of a head followed by exactly $2$ tails.

Let $Y$ be the # of sequences of a head followed by at least $2$ tails.

What are the values of $E[X]$ and $E[Y]$ ?

My efforts are meeting dead ends.

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@Ross Millikan: Done.Thank you for the correction. –  Ravi Donepudi Aug 12 '12 at 4:59
    
What do you mean by "# of sequences of a head followed by exactly 2 tails". Do you mean that there must be an occurence of "HTT" in the sequence, or does it have to start with "HTT"? –  utdiscant Aug 12 '12 at 7:31
    
@utdiscant: One or more HTT in the sequence. Needn't start with HTT. –  true blue anil Aug 12 '12 at 8:15

1 Answer 1

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  1. The probability that HTTH will appear in a particular position is $\dfrac{16 \choose 6}{20 \choose 8} = \dfrac{308}{4845}$ and there are 17 possible positions so $$E[X] = \frac{308}{285} \approx 1.0807.$$

  2. The probability that HTT will appear in a particular position is $\dfrac{17 \choose 7}{20 \choose 8} = \dfrac{44}{285}$ and there are 18 possible positions so $$E[Y] = \frac{264}{95} \approx 2.7789.$$

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Thanks, I was lost trying to compute the # of cases with 1,2... favorable sequences when only the expectation was asked for ! –  true blue anil Aug 12 '12 at 15:08

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