91Ó°ÊÓ

When coin 1 is flipped it lands on heads with probability of 0.4 when coin 2 is flipped it lands on heads with probability of 0.7 One of these coins is randomly chosen and flipped 10 times. Firstly let us calculate the probability of coin 1 and coin 2 landing exactly 7 times on heads.

$\mathrm{ℙ}(X=7)=\left(\begin{array}{c}10\\ 7\end{array}\right)·0.4^7·0.6^3=0.04$

$\mathrm{ℙ}(Y=7)=\left(\begin{array}{c}10\\ 7\end{array}\right)·0.7^7·0.3^3=0.27$

$\mathrm{â„™}\left(A\right)=\mathrm{â„™}(Aâˆ\pounds B=1)\mathrm{â„™}(B=1)+\mathrm{â„™}(Aâˆ\pounds B=2)\mathrm{â„™}(B=2)$

$=\frac{1}{2}·\mathrm{ℙ}(X=7)+\frac{1}{2}·\mathrm{ℙ}(Y=7)$

$=\frac{0.31}{2}=0.155$

The answer is$P\left(A\right)=$$0.155$$0.155$

In the (b) part of the assignment we want to calculate the probability that 7 out of 10 flips land on heads given that the first flips lands on heads. Let C be an event that first flip lands heads.

$\mathrm{â„™}(Aâˆ\pounds C)=\frac{\mathrm{â„™}(A,C)}{\mathrm{â„™}\left(C\right)}=\frac{\frac{1}{2}·\left(\begin{array}{l}9\\ 6\end{array}\right)·0.4^7·0.6^3+\frac{1}{2}·\left(\begin{array}{l}9\\ 6\end{array}\right)·0.7^7·0.3^3}{\frac{1}{2}·0.4+\frac{1}{2}·0.7}$

$=0.197$

The answer is$P\left(A\backslash C\right)=$$0.197$