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.

-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) = = 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

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:

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

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:

the ratio of white tiles to blue tiles is 48: 16. the ratio in it's simplest form is 3: 1.

step-by-step explanation:

ratio: take the first number and add a colon next to it, then add the second number after that. make sure the ratio order matches up with the written problem!

simplest form: we can simplify the ratio 48 : 16 by dividing both terms by the greatest common factor (gcf).

the gcf of 48 and 16 is 16.

divide both terms by 16:

48 ÷ 16 = 3

16 ÷ 16 = 1

reducing to simplest form:

48 : 16 = 3 : 1

step-by-step explanation:

goog be able to you with that maybe