Question

On three rolls of a single​ die, you will lose ​$19 if a 3 turns up at least​ once, and you will win ​$5 otherwise. What is the expected value of the​ game?
Let X be the random variable for the amount won on a single play of this game.

Answer 1

Answer:

-16 cents

Step-by-step explanation:

We are given that on  three rolls of a single​ die, you will lose ​$19 if a 3 turns up at least​ once, and you will win ​$5 otherwise.

We are to find the expected value of the game.

P (at least one 5 in three rolls) = 1 - P (no. of 3 in three) = $1-(\frac{3}{6} )^2$ = 0.875

P (other results) = 1 - 0.875 = 0.125

Random game value = -19, +5

Probabilities: 0.875, 0.125

Expected game value (X) = 0.875 × (-19) + 0.125 × (5) = -16 cents

Therefore, every time you play the game, you can expect to lose 16 cents

Answer 2

Answer:

It is expected to lose 5.10 dollars

Step-by-step explanation:

The probability of getting a 3 by throwing a die once is 1/6.

By throwing it 3 times the probability of not getting a 3 is:

$P=(\frac{5}{6}) ^ 3 =0.5787$

Then the probability of obtaining a three at least once in the 3 attempts is:

$P'=(1-0.5787)=0.421$

So if X is the discrete random variable that represents the amount gained in a single move of this game, the expected gain E(X) is:

$E(X)=P'*(X') + P*(X)$

$E(X) =0.421'*(-19) + 0.5787*(5)\\\\E(X) =-\ 5.10$