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If the probability of heads for coin 1 is 0.4, and the probability of heads is 0.7 for coin 2, and assuming either coin is selected with equal probability:

a. What is the probability that 7 of 10 flips is heads?

I did part a simply enough, $0.5[\left( \begin{matrix} 10\\ 7\end{matrix} \right) *0.4^7*0.6^3 + \left( \begin{matrix} 10\\ 7\end{matrix} \right) *0.7^7*0.3^3]$

Part b is is a bit confusing to me: What is the conditional probability that 7 of 10 flips is heads given that the 1st flip is heads?

I have event A: the first flip is heads, and event B, 7 of the 10 flips are heads:

So the $P(B|A)=\dfrac {P(A \cap B)} {P(A)}$. $P(A) = .5(0.4+0.7)=0.55$ How do I take the dependence of event B on event A into account when figuring this out? Our professor assured us that the answer in the solutions manual is wrong for part b.

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    $\begingroup$ I think you can use Bayes' theorem. $\endgroup$ Oct 29, 2015 at 4:06

2 Answers 2

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To calculate the probability of $A\cap B$, proceed more or less like in your calculation for a). With probability $0.5$ we are flipping the first coin. For that coin, calculate the probability $p_1$ that the first toss is head and that there are $6$ heads in the next $9$ tosses. This is $0.4$ times a "binomial" expression of a kind familiar to you. Define $p_2$ similarly. Then $\Pr(A\cap B)=(0.5)(p_1+p_2)$.

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  • Let $A$ be the event that: the 0.4 biased coin is selected.
  • Let $B$ be the event that: the 0.7 biased coin is selected.
  • Let $C$ be the event that: the first flip is a head
  • Let $D$ be the event that: the subsequent 9 flips contain 6 heads.

    • Noting that C and D will be conditionally independent given A (or B) we have:

$$\mathsf P(D\mid C) = \mathsf P(D\mid A)\;\mathsf P(A\mid C)+\mathsf P(D\mid B)\;\mathsf P(B\mid C)$$

Can you evaluate the RHS 's four conditional probabilities?

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