Answer to Question #349125 in Statistics and Probability for jaz

Question

Four cards are drawn from a desk without replacement. Find these probabilities.

a. All cards are jacks.

b. All cards are black cards.

c. All cards are hearts.

Answer 1

The probability formula states:

$P(A)=\frac{\text{number of favorable events}}{\text{number of total events}}.$

There are 52 cards in a deck, so there are

$C_{52}^4=\frac{52!}{4!\cdot(52-4)!}=\frac{52}{4!48!}=\frac{52\cdot51\cdot50\cdot49}{4\cdot3\cdot2\cdot1}=270,725$

ways for getting 4 cards.

a. There are 4 jacks in a deck, so there is

$C_4^4=\frac{4!}{4!\cdot(4-4)!}=\frac{4!}{4!0!}=1$

way to get all of them.

$P(\text{All cards are jacks})=\frac{C_4^4}{C_{52}^4}=\frac{1}{270,725}\approx$ 0.0000037.

b. There are 26 black cards in a deck (13 clubs and 13 spades), so there are

$C_{13}^4=\frac{26!}{4!\cdot(26-4)!}=\frac{26}{4!22!}=\frac{26\cdot25\cdot24\cdot23}{4\cdot3\cdot2\cdot1}=14,950$

ways to get 4 of them.

$P(\text{All cards are black cards})=\frac{C_{26}^4}{C_{52}^4}=\frac{14,950}{270,725}\approx$ 0.055222.

c. There are 13 hearts in a deck, so there are

$C_{13}^4=\frac{13!}{4!\cdot(13-4)!}=\frac{13!}{4!9!}=\frac{13\cdot12\cdot11\cdot10}{4\cdot3\cdot2\cdot1}=715$

ways to get 4 of them.

$P(\text{All cards are hearts})=\frac{C_{13}^4}{C_{52}^4}=\frac{715}{270,725}\approx$ 0.002641.

Answer. a. $\frac{C_4^4}{C_{52}^4}\approx$ 0.0000037

b. $\frac{C_{26}^4}{C_{52}^4}\approx$ 0.055222

c. $\frac{C_{13}^4}{C_{52}^4}\approx$ 0.002641