A deck of cards contains 52 cards, of which 4 are aces. You are offered the following wager: Draw one card at random from the deck. You win $10 if the card drawn is an ace. Otherwise, you lose $1. If you make this wager very many times, what will be the mean amount you win?

Using the principle of discrete probability, the expected value, which a measure of the mean amount after many plays is - 2/13

Calculating the required probabilities :

P(winning) = P(drawing an Ace) = 4/52 = 1/13

Hence, P(losing) = 1 - 1/13 = 12/13

X : 10 - 1

P(X) : _ 1/13 12/13

The expected value :

E(X) = 10 × (1/13) + - 1(12/13)

E(X) = 10/13 - 12/13

E(X) = - 2/13

Hence, the measure of the mean amount is -2/13.

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-$0.15

Step-by-step explanation:

We have been given win = $10, and loss = $-1

This deck of cards has 4 aces out of a total of 52 cards.

Then the probability can be explained as the number of favorable outcome/number of what is the possible outcome.

P(win) = number of favorable outcome/number of possible outcome

= 4/52

= 1/13

P(loss) = number of favorable outcome/number of possible outcome

= 48/52

= 12/13

Then we have

$10x1/13 + (-1)x12/13

= 10/13-12/13

= -2/13 dollars

= -$0.15

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step-by-step explanation: