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Question:

An unfair coin is flipped four times. $P(Heads)=0.6$ and $P(Tails)=0.4$. Every flip is independent of every other flip.
Find the probability of getting exactly $2$ heads and $2$ tails given that the first coin flip was a head.

For some reason I cannot seem to get the right answer. What I have tried:

$$P(A∩B) = 216/625$$ The above was obtained using $(nCx) p^x q^{n-x}$
$$ P(\text {Getting heads on first flip})=P(B) = 0.6 = 3/5$$ $$ P(A|B) = \frac {P(A∩B)}{P(B)}$$

I can't figure out where I am going wrong. Feeling dumb. Please help

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  • $\begingroup$ If someone could also let me know how to format the exponent correctly, I cannot seem to figure out how to add multiple values to q's exponent. Thanks :) $\endgroup$ Commented Aug 1, 2022 at 9:39
  • $\begingroup$ Use q^{n-x} to get $q^{n-x}$. $\endgroup$
    – Cathedral
    Commented Aug 1, 2022 at 9:41
  • $\begingroup$ @Cathedral Thank you very much! $\endgroup$ Commented Aug 1, 2022 at 9:42
  • $\begingroup$ Given the initial $H$, the question comes down to "what's the probability of getting exactly one $H$ out of $3$?" which is a straight binomial computation. $\endgroup$
    – lulu
    Commented Aug 1, 2022 at 9:50
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    $\begingroup$ @Cathedral Yea it does. That was the first method I used, I have spent about 40 minutes trying to figure out where I was going wrong. Thanks :D $\endgroup$ Commented Aug 1, 2022 at 9:59

1 Answer 1

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Let $X_1,X_2,X_3,X_4$ be i.i.d. random variables with $\mathbb P(X_1=1)=p=1-\mathbb P(X_1=0)$. We compute the conditional probability $$ \mathbb P\left(\sum_{i=1}^4 X_i=2\mid X_1=1\right) = \mathbb P\left(\sum_{i=2}^4 X_i=1\right) = 3 p (1-p)^2. $$ With $p=\frac35$, this reduces to $\frac{36}{125}$.

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