Lecture 9, Problems 9.9 a) and b)

For 9.9 a, I got $\dfrac{3}{50}$ because you can separate the problem into two cases. The first case is where you get tails, and the second case is where you get heads. The probability you get tails on the die is $\dfrac{1}{2}$, and the probability you make given that you got tails is $\dfrac{1}{25}$. The product is $\dfrac{1}{50}$. In the second case, the probability you get heads is $\dfrac{1}{2}$, and the probability you make can be separate into two subcases. One is where you make on your first try, and the second is when you make on the second try. The probability you make on the first try is $\dfrac{1}{25}$. The probability you make on the second try is $\dfrac{24}{25} \cdot \dfrac{1}{25}$. The probability for heads should be $\dfrac{2}{25}$. The answer for the problem should be $\dfrac{3}{50}$, but the answer in the book says that it's $\dfrac{1}{2} \cdot \dfrac{1}{25} + \dfrac{1}{2} \cdot \left[1-\left(\dfrac{24}{25}\right)\right]$.

For 9.9 b, I got $\dfrac{2}{3}$ because the probability is equal to the probability you get heads and win divided by the probability you win. The answer should be $\dfrac{2}{3}$, but the answer in the book says that it's $\dfrac{\dfrac{1}{2}}{\dfrac{1}{2} \cdot \dfrac{1}{25} + \dfrac{1}{2} \cdot \left[1-\left(\dfrac{24}{25}\right)\right]}$.