Question

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

Answer 1

Answer:

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