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.

P ( E_1*E_2*E_3*E_4 ) = 0.1055

Step-by-step explanation:

Given:

- 52 cards are dealt in 1 , 2 , 3 , 4 hands.

- Events:

             E_1       Hand 1 has exactly 1 ace

             E_2       Hand 2 has exactly 1 ace

             E_3       Hand 3 has exactly 1 ace

             E_4       Hand 4 has exactly 1 ace

Find:

p =P ( E_1*E_2*E_3*E_4 )

Solution:

Multiplication rule.

- For n ε N and events E_1 , E_2 , ... , E_n:

P ( E_1*E_2**E_n ) = P (E_1)*P(E_2|E_1)*P(E_3|E_2*E_1)**(E_n|E_1*E_2...E_n-1 )

- So for these events calculate 4 probabilities:-

-  For E_1, is to choose an ace multiplied by the number of ways to choose remaining 12 cards out of 48 non-aces:

                               P ( E_1 ) = 4C1 * 48C12 / 52C13

- For E_2 | E_1 , one ace and 12 other cards have already been chosen. there are 39C13 equally likely hands. The number of different one ace hand 2 is the number of ways to choose an ace from 3 remaining multiplied by the number of ways to choose the remaining 12 from 36, we have:

                               P ( E_2 | E_1  ) = 3C1 * 36C12 / 39C13

                               P ( E_3| E_2*E_1  ) = 2C1 * 24C12 / 26C13

                               P ( E_4 | E_3*E_2*E_1  ) = 1C1*12C12 / 13C13 = 1

- So the multiplication rule for n = 4 is as follows:

     P ( E_1*E_2*E_3*E_4 ) = P (E_1)*P(E_2|E_1)*P(E_3|E_2*E_1)*P ( E_4 | E_3*E_2*E_1  ) = [ 4C1 * 48C12 / 52C13 ] * [ 3C1 * 36C12 / 39C13 ] * [ 2C1 * 24C12 / 26C13 ]

     P ( E_1*E_2*E_3*E_4 ) = [ 4!*48! / (12!)^4 ] / [ 52! / (13!)^4 ]

     P ( E_1*E_2*E_3*E_4 ) = [ 4!*13^4 / (52*51*50*49) ]

    P ( E_1*E_2*E_3*E_4 ) = 0.1055

yes

\

step-by-step explanation: