91Ó°ÊÓ

Given in the question that, here are two misshapen coins in a box; their probabilities for landing on heads when they are flipped are, respectively, .$4$and .7. One of the coins is to be randomly chosen and flipped $10$times. Given that two of the first three flips landed on heads

Assuming we are given that $Y=5$, that truly intends that in fifth throw we have gotten five and that there is no fives inside initial four throws. Utilizing the law of the absolute assumption, we have that

$E_C\left(X\right)=2+E_C\left(Y\right)$

Utilizing the law of the complete probability, we have that

$E_C\left(Y\right)=E_C(Yâˆ\pounds A)P_C\left(A\right)+E_C(Yâˆ\pounds B)P_C\left(B\right)$

We should decide $P_C\left(A\right)$. Utilizing the Bayesian formula, we have that

$P_C\left(A\right)=P(Aâˆ\pounds C)=\frac{P(Câˆ\pounds A)P\left(A\right)}{P\left(C\right)}=\frac{P(Câˆ\pounds A)P\left(A\right)}{P(Câˆ\pounds A)P\left(A\right)+P(Câˆ\pounds B)P\left(B\right)}$

$=\frac{\left(\begin{array}{l}3\\ 2\end{array}\right)·0.4^2·0.6·0.5}{\left(\begin{array}{l}3\\ 2\end{array}\right)·0.4^2·0.6·0.5+\left(\begin{array}{l}3\\ 2\end{array}\right)·0.7^2·0.3·0.5}=\frac{32}{81}$

That implies that

$P_C\left(B\right)=1-P_C\left(A\right)=\frac{49}{81}$

Finally, we have that

$E_C(Yâˆ\pounds A)=E(Yâˆ\pounds A)=7·0.4=2.8$

$E_C(Yâˆ\pounds B)=E(Yâˆ\pounds B)=7·0.7=4.9$

which yields

$E_C\left(X\right)=\frac{1639}{270}≈6.0704$

The required mean is$E_C\left(X\right)≈6.0704$