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This is taken from Ross book: when coin $1$ is flipped, it lands on heads with probability .4; when coin $2$ is flipped, it lands on heads with probability $.7$. One of these coins is randomly chosen and flipped $10$ times. Compute the probability that the coin lands on heads on exactly $7$ of the $10$ flips.

What is incorrect behind this reasoning? let $C$ be the event that the coin lands on heads. By total probability we have $P(C)=\frac{1}{2}.7+\frac{1}{2}.4=.55$. Then we apply a binomial distribution with $n=10$, $k=7$ and $p=.55$. However, this is incorrect. Why?

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  • $\begingroup$ It may just be as ambiguity in the statement. I think he means "First choose a coin, then flip the chosen coin ten times." You are interpreting it to mean, "Ten times in a row, choose a coin a flip it it." I would have taken it to mean the first thing, but it seems to bear the second interpretation just as well. $\endgroup$ – saulspatz Feb 6 '18 at 20:47
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You don't have the event $C$, you have ten events $C_1,\ldots,C_{10}$ where $C_j$ is the event of the $j$-th toss being heads. You are assuming these events are independent, but they are not. Essentially this is because it's the same coin involved in any pair of tosses. In detail the probability of $C_1\cap C_2$ is $\frac12(0.4^2+0.7^2)$ which isn't the square of $P(C_i)=\frac12(0.4+0.7)$.

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You have to condition first on the coin chosen, then calculate the probabilistic density function for each coin.

Let $C$ be the event that coin $1$ is chosen, and let $X$ be the number of heads.

\begin{align} & P(X = 7) \\ =& P(X = 7 | C = 1) P(C = 1) + P(X = 7 | C = 2) P(C = 2) \\ =& {10 \choose 7}(0.4)^7(1-0.4)^3 \cdot \frac12 + {10 \choose 7}(0.7)^7(1-0.7)^3 \cdot \frac12 \\ =& \frac{3399}{80038} + \frac{2537}{9508} \\ =& \frac{5091}{16460} \end{align}

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