Question

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.

(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?

Short answer

The answer for part (a) is$0.155$and part (b) is$0.197$

Step-by-step solution

Step 1:Given Information (Part-a)

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.

Step 2:Calculation (Part-a)

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

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

$\mathrm{\mathbb{P}}\left(A\right)=\mathrm{\mathbb{P}}(A\mid B=1)\mathrm{\mathbb{P}}(B=1)+\mathrm{\mathbb{P}}(A\mid B=2)\mathrm{\mathbb{P}}(B=2)$

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

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

Step 3:Final Answer (Part-a)

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

Step 4:Given Information (Part-b)

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.

Step 5:Calculation(Part-b)

$\mathrm{\mathbb{P}}(A\mid C)=\frac{\mathrm{\mathbb{P}}(A,C)}{\mathrm{\mathbb{P}}\left(C\right)}=\frac{\frac{1}{2}·\left(\begin{array}{l}9\\ 6\end{array}\right)·0.4^7·0.6^3+\frac{1}{2}·\left(\begin{array}{l}9\\ 6\end{array}\right)·0.7^7·0.3^3}{\frac{1}{2}·0.4+\frac{1}{2}·0.7}$

$=0.197$

Step 6:Final Answer(Part-b)

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