Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining p, the probability that each hand has an ace. Let Ei be the event that the ith hand has exactly one ace. Determine p = P(E1E2E3E4) by using the multiplication rule.

Solution:

Let  E1 = be the event of the first hand of 13 cards having exactly one ace. 

The probability that the first hand will have exactly one ace denoted by p(E₁) is

p(E₁) = 4/52 = 1/13      

When the first hand is randomly selected the number of aces in the pack of 52 cards is 4.

The probability that the second hand of 13 cards will have one ace denoted by p(E₂)  is given by:

p(E₂) = 3/39 = 1/13

Since one ace is already chosen in the first hand there are only three aces left in total 39 cards left.

The probability that the third hand of 13 cards will have one ace denoted by p(E₃)  is given by:

p(E₃) = 2/26 = 1/13

Since one ace is already chosen in the first hand there are only two aces left in total of 26 cards left.

The probability that the fourth hand of 13 cards will have one ace denoted by p(E₄)  is given by:

p(E₄) = 1/13 = 1/13

Since one ace is already chosen  in the first hand there are only one ace left in a total of 13 cards left.

The probability that the four hands (0f 13 cards each) will have one ace each is given by multiplication of the four probabilities

p(E₁E₂E₃E₄) = p(E₁) × p(E₂) × p(E₃) × p(E₄)  = (1/13)(1/13)(1/13)(1/13) = 1/28561

Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining p, the probability that each hand has an ace. Let Ei be the event that the ith hand has exactly one ace. Determine p = P(E1E2E3E4) by using the multiplication rule.

Summary:

If  an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each, the probability that each hand has an ace will be 1/13 and the probability that all four hands will have one ace each  p = P(E1E2E3E4) = 1/28561