Question

1. use the picture of all 52 cards in a standard deck below to answer the questions. you are drawing random 3 - card hands for all of these problems.

a. how many possible 3 - card hands are there in a deck of 52 cards?
b. what is the probability that all 3 cards are red?
c. what is the probability that all three cards are spades?
d. what is the probability that you draw exactly 1 ace?
e. what is the probability that you draw two red cards and one black card?
f. what is the probability that you draw two black cards and one red card?
g. what is the probability that you draw two cards of one color and one card of the other color?

Explanation:

Step1: Calculate total 3 - card hands

The number of ways to choose 3 cards from 52 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 52$ and $r=3$. So $C(52,3)=\frac{52!}{3!(52 - 3)!}=\frac{52\times51\times50}{3\times2\times1}=22100$.

Step2: Calculate probability of all - red hands

There are 26 red cards. The number of ways to choose 3 red cards from 26 is $C(26,3)=\frac{26!}{3!(26 - 3)!}=\frac{26\times25\times24}{3\times2\times1}=2600$. The probability $P=\frac{C(26,3)}{C(52,3)}=\frac{2600}{22100}=\frac{2600\div100}{22100\div100}=\frac{26}{221}=\frac{2}{17}$.

Step3: Calculate probability of all - spades hands

There are 13 spades. The number of ways to choose 3 spades from 13 is $C(13,3)=\frac{13!}{3!(13 - 3)!}=\frac{13\times12\times11}{3\times2\times1}=286$. The probability $P=\frac{C(13,3)}{C(52,3)}=\frac{286}{22100}=\frac{11}{850}$.

Step4: Calculate probability of exactly 1 ace

There are 4 aces and 48 non - aces. The number of ways to choose 1 ace from 4 and 2 non - aces from 48 is $C(4,1)\times C(48,2)$. $C(4,1)=\frac{4!}{1!(4 - 1)!}=4$, $C(48,2)=\frac{48!}{2!(48 - 2)!}=\frac{48\times47}{2\times1}=1128$. So $C(4,1)\times C(48,2)=4\times1128 = 4512$. The probability $P=\frac{C(4,1)\times C(48,2)}{C(52,3)}=\frac{4512}{22100}=\frac{1128}{5525}$.

Step5: Calculate probability of 2 red and 1 black

There are 26 red and 26 black cards. The number of ways to choose 2 red cards from 26 and 1 black card from 26 is $C(26,2)\times C(26,1)$. $C(26,2)=\frac{26!}{2!(26 - 2)!}=\frac{26\times25}{2\times1}=325$, $C(26,1)=26$. So $C(26,2)\times C(26,1)=325\times26 = 8450$. The probability $P=\frac{C(26,2)\times C(26,1)}{C(52,3)}=\frac{8450}{22100}=\frac{65}{170}=\frac{13}{34}$.

Step6: Calculate probability of 2 black and 1 red

The number of ways to choose 2 black cards from 26 and 1 red card from 26 is $C(26,2)\times C(26,1)$ (same calculation as step 5). So the probability is also $\frac{13}{34}$.

Step7: Calculate probability of 2 of one color and 1 of the other

Since the probability of 2 red and 1 black is the same as 2 black and 1 red, the probability is $\frac{13}{34}+\frac{13}{34}=\frac{13}{17}$.

Answer:

a. 22100
b. $\frac{2}{17}$
c. $\frac{11}{850}$
d. $\frac{1128}{5525}$
e. $\frac{13}{34}$
f. $\frac{13}{34}$
g. $\frac{13}{17}$