Compute the expectation of X When a certain basketball player takes his first shot in a game with probability 1/3.
If he misses his first 2 shots, his third shot will go with probability 1/5.
If he misses his first 3 shots, his fourth shot will go with prob. 1/6.
If he misses his first 4 shots, the coach will remove him from the game.
Assume that the player keeps shooting until he succeeds or he is removed from the game. Let X denote the number of shots he misses until his first success or until he is removed from the game.
Compute $\mathbb{E}[X]$
If I separate this into indicator functions then the expectation is the sum of the probabilities of each event happening. This would mean thst if I find the pmf and then sum the probabilities, I would get the expectation. My concern is that this would mean the expectation is $1$. Is that a reasonable expectation? The pmf I found is as follows:
$$\mathbb{P}(X=1)=\mathbb{P}(\text{misses 1st and makes second})=\frac{2}{3}*\frac{1}{4}$$
$$\mathbb{P}(X=2)=\mathbb{P}(\text{misses 1st, 2nd and makes third})=\frac{2}{3}*\frac{3}{4}*\frac{1}{5}$$
$$\mathbb{P}(X=3)=\mathbb{P}(\text{misses 1st,2nd,3rd and makes fourth})=\frac{2}{3}*\frac{3}{4}*\frac{4}{5}*\frac{1}{6}$$
$$\mathbb{P}(X=4)=\mathbb{P}(\text{misses 1st,2nd,3rd,4th)}=\frac{2}{3}*\frac{3}{4}*\frac{4}{5}*\frac{5}{6}$$
All of this sums to $1$ as a pmf should.
Edit: I forgot that there is also the probability that he misses 0 shots. So I should add $\mathbb{P}(X=0)=\frac{1}{3}$
Now it should add up to 1.
 A: You don't use an indicator r.v. here, what you need to do is to compute $\Bbb E[X] = \Sigma\left([X_i]\cdot \Bbb P[i]\right)$
so $\Bbb E[X] = 0\cdot\Bbb P[0] + 1\cdot\Bbb P[1] + ... 4\cdot\Bbb P[4]$
= $0*1/3 +1*1/6 + 2*1/10 + 3*1/15 + 4*1/3 = 1.9$
A: Let X denote the number of shots he misses until his first success or until he is removed from the game.
Then, X can be ${0, 1, 2, 3, 4}$
The probability that $X = 0$ is $1/2$. i.e $P(X = 0 ) = 1/2 = 0.5$
$P (X = 1) = P$ (Misses 1st shot AND Makes 2nd shot) =$P$ (Misses 1st shot). $P$(Makes 2nd shot | Misses 1st shot)  $=(\frac{1}{2}).(\frac{1}{3}) = 0.167$
$P (X = 2) = P$ (Misses 1st shot AND Misses 2nd shot AND Makes 3rd shot) =$P$ (Misses 1st shot AND 2nd shot). $P$(Makes 3rd shot | Misses 1st shot AND 2nd shot) $ = \frac{1}{2}.(1 - \frac{1}{3}).\frac{1}{4} = (\frac{1}{2}).(\frac{2}{3}).(\frac{1}{4}) = 0.0833$
$P (X = 3) = (\frac{1}{2}).(\frac{2}{3}).(1 - \frac{1}{4}).( \frac{1}{5}) =(\frac{1}{2}).(\frac{2}{3}).( \frac{3}{4}).( \frac{1}{5}) = 0.05 $
If he misses his first 4 shots then the coach will remove him from the game.
Therefore,
$P (X = 4) = (\frac{1}{2}).(\frac{2}{3}).( \frac{3}{4}).( \frac{1}{5}) = .20 $
Then, PMF = $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.5+ 0.167 + 0.0833 + 0.05+0.20 = 1  $
