Question

Jordan wants to play a basketball game at a carnival. The game costs the player $5 dollar sign, 5 to play, and the player gets to take two long-distance shots. If they miss both shots, they get nothing. If they make one shot, they get their $5 dollar sign, 5 back. If they make both shots, they get $10 dollar sign, 10 back. Jordan has a 40% percent chance of making this type of shot.
Here is the probability distribution of X equals the number of shots Jordan makes in a randomly selected game, and equals the amount of money Jordan gains from playing the game.
X=\# \text{ of shots made}X=# of shots madeX, equals, \#, start text, space, o, f, space, s, h, o, t, s, space, m, a, d, e, end text 000 111 222 M=\text{money gained}M=money gainedM, equals, start text, m, o, n, e, y, space, g, a, i, n, e, d, end text -\$5−$5minus, dollar sign, 5 \$0$0dollar sign, 0 \$5$5dollar sign, 5 Probability 0.360.360, point, 36 0.480.480, point, 48 0.160.160, point, 16
Find the expected value of the amount of money Jordan gains from playing this game.

Answer 1

Answer:

The expected value of Jordan gains is -1 dollar.

Step-by-step explanation:

Consider the following random variables. X := #of shots that Jordan makes. Then, we can can define the random variable Y of the earnings of Jordan in a game as follows

Y = 5 if X=2 (since he gets 10, but invested 5),  Y=0 if X=1(since he gets the 5 back) and Y=-5 if X=0(since he doesn't get the money back). Then, in this case, we can define the probability as follows.

P(Y=5) = P(X=2), P(Y=0) = P(X=1), P(Y=-5)= P(X=0).

By definition, the expected value of Y is given by

$E[Y] = 5\cdot P(Y=5)+0\cdot P(Y=0)-5 P(Y=-5)$. By the previous analysis, we have that

$E[Y] = 5\cdot P(X=2)-5P(X=0)$

We only need to calculate the probabilities for X. In this case, we can consider each shot independt from each other. Then, we can consider X to be distributed as a binomial random variable with n=2 trials and p=0.4 of success (since he has a 40% chance of winning).

Then, by definition

$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k} = \binom{2}{k}0.4^{k}0.6^{2-k}$

where $\binom{n}{k}=\frac{n!}{k!(n-k)!}$

Then,

$P(X=0) = \frac{2!}{0!2!}0.4^{0}0.6^{2} = 0.36$

$P(X=2)=\frac{2!}{0!2!}0.4^{2}0.6^{0} = 0.16$

Then,

$E[Y] = 5\cdot 0.16-5\cdot 0.36 = -1$