AMDM Question (Mathematics)

Question

In the playing card deck below what is the chance of pulling 4 face cards without replacing the cards in between pulls? Answer in decimal form rounded to the 6th digit after the decimal

Answer 1

In the playing card deck below what is the chance of pulling 4 face cards without replacing the cards in between pulls? Answer in decimal form rounded to the 6th digit after the decimal

Detailed Solution:

============

Let face cards be: F1, F2, F3, F4

Total Number of Cards: 52 (Decreases by 1 after each pulling)

Total Number of Face Cards: 12 (Decreases by 1 after each pulling)

In 1st pulling - Number of Face Cards: 12 ==> P(F1) = 12/52 = 3/13

In 2nd pulling - Number of Face Cards: 11 ==> P(F2) = 11/51

In 3rd pulling - Number of Face Cards: 10 ==> P(F3) = 10/50 = 1/5

In 4th pulling - Number of Face Cards: 9 ==> P(F4) = 9/49

P(F) = P(F1)*P(F2)*P(F3)*P(F4) = (3/13)*(11/51)*(1/5)*(9/49) = (297)/(162435) = 0.0018284237

P(F) = 0.001828 (Chance of pulling 4 face cards without replacing the cards in between pulls - rounded to the 6th digit)

Answer 2

There are 3 face cards per suit and 4 suits yielding 12 face cards. There are 52 total cards. So, in draw #1 there is a probability of 12/52 of drawing a face card.

In draw #2, you have already drawn a face card so there are now 11 face cards and 51 total cards making the probability 11/51

In draw #3, the probability of drawing another face card is 10/50

In draw # 4, it is 9/49

The probability that all those events occurred is 12/52 • 11/51 • 10/50 • 9/49 = 11880/6497400 = 99/54145 = 0.001828