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When coin $1$ is flipped, it lands on heads with probability $0.4$; when coin $2$ is flipped, it lands on heads with probability $0.7$. One of these coins is randomly chosen and flipped $10$ times.

(a) What is the probability that the coin lands on heads on exactly $7$ of the $10$ flips?

(b) Given that the first of these $10$ flips lands heads, what is the conditional probability that exactly $7$ of the $10$ flips land on heads?

MY ATTEMPT

(a) Let us denote by $X$ the random variable which counts the numbers of heads in 10 flips. Thus $X\sim\text{Binomial}(10,p)$, where the parameter $p$ depends on the coin that is chosen: $C = 1$ stands for the event "the first coin is chosen" and $C = 2$ stands for the event "the second coin is chosen". Based on these considerations, we have the answer to the first question, which is

\begin{align*} \textbf{P}(X = 7) = \textbf{P}(X = 7\mid C = 1)\textbf{P}(C = 1) + \textbf{P}(X = 7\mid C = 2)\textbf{P}(C = 2) \end{align*}

where \begin{cases} \textbf{P}(X = 7\mid C = 1) = \displaystyle{10\choose 7}(0.4)^{7}(0,6)^{3}\\\\ \textbf{P}(X = 7\mid C = 2) = \displaystyle{10\choose 7}(0.7)^{7}(0,3)^{3}\\\\ \textbf{P}(C = 1) = \textbf{P}(C = 2) = 0.5 \end{cases}

(b) Since $X = \displaystyle\sum_{k=1}^{10} X_{k}$, where each $X_{k}\sim\text{Bernoulli}(p)$, we are interested in the event $\textbf{P}(X = 7\mid X_{1} = 1)$, which is equivalent to

\begin{align*} \textbf{P}(X = 7\mid X_{1} = 1) & = \frac{\textbf{P}(X = 7, X_{1} = 1)}{\textbf{P}(X_{1} = 1)} = \frac{\textbf{P}\left(\displaystyle\sum_{k=2}^{10}X_{k} = 6\right)}{\textbf{P}(X_{1} = 1)} \end{align*}

Since $\textbf{P}(X_{1} = 1) = \textbf{P}(X_{1} = 1 | C = 1)\textbf{P}(C = 1) + \textbf{P}(X_{1} = 1 | C = 2)\textbf{P}(C = 2)$, it suffices to calculate $\textbf{P}(Y = 6)$, where $Y := \displaystyle\sum_{k=2}^{10}X_{k}$ and, consequentely, $Y\sim\text{Binomial}(9,p)$:

\begin{align*} \textbf{P}(Y = 6) = \textbf{P}(Y = 6\mid C = 1)\textbf{P}(C = 1) + \textbf{P}(Y = 6\mid C = 2)\textbf{P}(C = 2) \end{align*}

where \begin{cases} \textbf{P}(Y = 6\mid C = 1) = \displaystyle{9\choose 6}(0.4)^{6}(0,6)^{3}\\\\ \textbf{P}(Y = 6\mid C = 2) = \displaystyle{9\choose 6}(0.7)^{6}(0,3)^{3}\\\\ \textbf{P}(C = 1) = \textbf{P}(C = 2) = 0.5 \end{cases}

My question is: am I working it right or is there any conceptual misapplication?

Any contribution is appreciated. Thanks in advance!

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For part $a)$,

Let $K_1$ denotes when coin $1$ is chosen and $K_2$ denotes when coin $2$ is chosen.

The probability that the coin lands on heads on exactly $7$ of the $10$ flips is $P(H)$ $$P(H)=P(H|K_1)P(K_1)+P(H|K_2)P(K_2)$$ $$P(H)=\dbinom{10}{7}(0.4)^7(0.6)^3(0.5)+\dbinom{10}{7}(0.7)^7(0.3)^3()0.5$$

For part $b)$,

When the first flip is head, then the probability is computed from choosing $6$ head flips out of $9$ flips in both coins case. So, we get $\dbinom{9}{6}(0.4)^6(0.6)^3(0.5)+\dbinom96(0.7)^6(0.3)^3(0.5)$

Let $T$ denotes the event that the first flip is head.

$$P(H|T)=\dfrac{P(H\cap T)}{P(T)}$$ $P(T)=P(T|K_1)P(K_1)+P(T|K_2)P(K_2)=(0.4)(0.5)+(0.7)(0.5)=0.55$

$P(H\cap T)=P(H\cap T|K_1)P(K_1)+P(H\cap T|K_2)P(K_2)$ $$=\dbinom96(0.4)^6(0.6)^3(0.5)+\dbinom96(0.7)^6(0.3)^3(0.5)$$

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  • $\begingroup$ In the first place, thanks for the contribution Flex! As far as I have checked, we have given the same answers. Maybe there is just one typo: it should be $(0,4)^{6}(0.6)^{3}(0.5)$ and $(0.7)^{6}(0.3)^{3}(0.5)$ in the last line. $\endgroup$ – user1337 Jan 27 '19 at 2:39
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    $\begingroup$ @user1337 Fixed it! Thanks $\endgroup$ – Key Flex Jan 27 '19 at 2:44

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