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Suppose we flip 10 times an unfair coin that fall a probability of $p$ on 'heads'. Knowing that we obtained 'heads' 6 times out of the 10 flips, find the conditionnal probability of the first three flips being heads, tails, tails.

My effort: Since we don't know the exact probability of getting heads i would think it would be somthing like:

A: Probability of getting 6 'head' out of 10 B: Probability for the first 3 flip to be 'heads''tails''tails'

$A:\binom{10}{6}(p)^6(1-p)^4$

$B:p(1-p)^2$

$(B|A):???$

Any help to point me in the right direction would be greatly appreciated.

Thank you.

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    $\begingroup$ Hint: The event $\{ B \cap A\}$ is equivalent to getting heads exactly five times in the next 7 tosses. $\endgroup$ – stochasticboy321 Nov 11 '15 at 1:17
  • $\begingroup$ So B|A would be $(\frac{1}{3})(1-B\cap A)$? $\endgroup$ – spexel Nov 11 '15 at 1:22
  • $\begingroup$ Nope. What doe $1 - A$ even mean of an event $A$? To calculate the probability, recall Bayes' theorem : $P(B|A) = \frac{P(B \cap A)}{P(A)}$ $\endgroup$ – stochasticboy321 Nov 11 '15 at 1:29
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$P(\text{first 3 flips were HTT }|\text{ 6/10 flips were heads overall})$

$=\frac{P(\text{first 3 flips were HTT } \cap \text{ 6/10 flips were heads overall}) }{ P(\text{6/10 flips were heads overall})}$

$=\frac{P(\text{first 3 flips were HTT } \cap \text{ 5 heads in the final 7 flips}) }{ P(\text{6/10 flips were heads overall})}$

$=\frac{p(1-p)(1-p) \cdot \binom{7}{5}p^5(1-p)^2}{ \binom{10}{6}p^6(1-p)^4}$

$=\frac{1}{10}$

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The condition you have is that there is some arrangement of six heads, and four tails.

All such arrangements of those ten results are equally probable.   We don't need to know what that probability is, just that all of them are identical.

What then is the conditional probability that the first three coins have the results: "heads, heads, tails," in that specific order, given that all ten coins present one of the above arrangements?

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