3.2.8 problem 24

(a)

Since tosses are independent, we expect information about two of the tosses to not provide any information about the remaining tosses. In other words, we expect the required probability to be

$$\left(\frac{8}{k}\right){(0.5)}^k{(0.5)}^{8-k}=\left(\frac{8}{k}\right){(0.5)}^8$$

for $0\le k\le 8$.

To prove this, let $X$ be the number of Heads out of the $10$ tosses, and let $X_{1,2}$ be the number of Heads out of the first two tosses.

$$\begin{array}{cccccccc}P(X=k|X_{1,2}=2)& =\frac{P(X=k\cap X_{1,2}=2)}{P(X_{1,2}=2)}& & & & & & \\ & =\frac{{(0.5)}^2\left(\frac{8}{k-2}\right){(0.5)}^{k-2}{(0.5)}^{8-k+2}}{{(0.5)}^2}& & & & & & \\ & =\left(\frac{8}{k-2}\right){(0.5)}^{k-2}{(0.5)}^{8-k+2}& & & & & & \\ & =\left(\frac{8}{k-2}\right){(0.5)}^8& & & & & & \end{array}$$

for $2\le k\le 10$, which is equivalent to $\left(\frac{8}{k}\right){(0.5)}^8$ for $0\le k\le 8$.

(b)

Let $X_{\ge 2}$ be the event that at least two tosses land Heads.

$$\begin{array}{cccccccc}P(X=k|X_{\ge 2})& =\frac{P(X=k\cap X_{\ge 2})}{X_{\ge 2}}& & & & & & \\ & =\frac{\left(\frac{10}{k}\right){(0.5)}^k{(0.5)}^{10-k}}{1-(0.5^{10}+10\ast 0.5^{10})}& & & & & & \end{array}$$

for $2\le k\le 10$.

To see that this answer makes sense, notice that if we over all values of $k$ from $2$ to $10$, we get exactly the denominator, which means the said sum equals to $1$.