Chapter 3 : 3.12 - Exercises Problem 7

[Sahil Khan]

The solution given by you here seems incorrect to me. There should be (1-px) term in the end to account for failing.

Refer - math.stackexchange

[vmdd]

I guess you meant this problem (or the link might have changed).

For the case of only reaching lvl 1 and no futher, we need to account for failure to go to lvl2 $(1 - p_1)$, so

$$P(X = 1) = 1 \cdot (1 - p_1)$$

Also, it cannot be equal $1$ at $X = 1$, since that would mean it needs to be $0$ at any other point.

[RamonAraujo]

I agree with @vmdd . From the question statement, I interpret that X is the highest level that Bob can reach by playing the game. X = 1 if Bob fails to pass to level 2, which occurs with probability (1-p1).

Therefore, I think the correct answer to question 3.12 is

$$P(X=k) = \begin{cases} 1-p_1 &, ~\text{for}~ k=1\\ p_1 p_2 \dots p_{k-1} (1-p_k) &, ~\text{for}~ k=\{ 2,3,4,5,6 \} \\ p_1 p_2 p_3 p_4 p_5 p_6 &, ~\text{for}~ k=7 \\ 0 &, ~\text{otherwise}~ \end{cases}$$