Suppose you first randomly sample one card from a deck of 52. Then, without putting the card back in the deck you sample a second and then (again without replacing cards) a third. Given this sampling procedure, what is the probability that at least one ace will be in the three sampled cards

P(A) = 1201/5525 or 0.2174

The probability that at least one ace will be in the three sampled cards is 1201/5525 or 0.2174

Step-by-step explanation:

Let P(A) represent the probability that at least one ace will be in the three sampled cards.

P(A) = 1 - P(A') 1

Where;

P(A') is the probability that none of the three cards will be ace.

Number of ace in a standard 52 cards = 4

Number of non-ace in a standard 52 cards = 52-4 =48

P(A') = 48/52 × 47/51 × 46/50

P(A') = 4324/5525 = 0.7826

Substituting, the value into equation 1;

P(A) = 1 - P(A')

P(A) = 1 - 4324/5525

P(A) = 1201/5525 or 0.2174

The probability that at least one ace will be in the three sampled cards is 1201/5525 or 0.2174

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00

21

jk it's 20

step-by-step explanation: