Question

stat 213 assignment 1: problem 8
(1 point)
you are to pick 4 cards randomly from a condensed deck of cards that contains four suits - ♣,♦,♥,♠ and the following denominations: ace,2,3,4,5,6,7,8,9 and 10. there are no face - cards in this deck.
what is the probability that you will get
part (a) a three - of - a - kind (three of the same denomination)?
p(3 of a kind)= (use four decimals)
part (b) what is the probability that all four cards are of the same suit? for example, all four cards are ♣s?
p(all the same suit)= (use four decimals)
part (c) what is the probability you get one pair?
p(one pair)= (use four decimals)
part (d) what is the probability you get two aces and two 10s?
p(2 - aces and 2 - 10s)= (use four decimals)
part (e) what is the probability you get two aces and two ♣s?
p(2 - aces and 2 - ♣)= (use four decimals)

Explanation:

Step1: Calculate total number of ways to pick 4 cards

The deck has $4\times9 = 36$ cards. The total number of ways to choose 4 cards out of 36 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 36$ and $r=4$. So, $C(36,4)=\frac{36!}{4!(36 - 4)!}=\frac{36\times35\times34\times33}{4\times3\times2\times1}=58905$.

Step2: Calculate probability for part (a)

• First, choose the denomination for the three - of - a - kind. There are 9 denominations, so $C(9,1) = 9$ ways to choose the denomination for the three - of - a - kind.
• For the chosen denomination, there are $C(4,3)$ ways to pick 3 cards out of 4 of that denomination, and $C(32,1)$ ways to pick the remaining 4th card. $C(4,3)=\frac{4!}{3!(4 - 3)!}=4$ and $C(32,1)=\frac{32!}{1!(32 - 1)!}=32$. The number of three - of - a - kind hands is $9\times4\times32 = 1152$.
• The probability $P(3\text{ of a kind})=\frac{1152}{58905}\approx0.0196$.

Step3: Calculate probability for part (b)

• Choose a suit, there are 4 suits, so $C(4,1) = 4$ ways to choose the suit.
• Then choose 4 cards out of 9 of that suit, $C(9,4)=\frac{9!}{4!(9 - 4)!}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126$. The number of hands with all cards of the same suit is $4\times126 = 504$.
• The probability $P(\text{all the same suit})=\frac{504}{58905}\approx0.0086$.

Step4: Calculate probability for part (c)

• Choose the denomination for the pair: $C(9,1)=9$ ways.
• Choose 2 cards out of 4 of that denomination: $C(4,2)=\frac{4!}{2!(4 - 2)!}=6$.
• Choose 2 different denominations from the remaining 8 denominations for the non - paired cards: $C(8,2)=\frac{8!}{2!(8 - 2)!}=28$.
• For each of the 2 chosen non - paired denominations, choose 1 card out of 4: $C(4,1)\times C(4,1)=4\times4 = 16$. The number of one - pair hands is $9\times6\times28\times16=24192$.
• The probability $P(\text{one pair})=\frac{24192}{58905}\approx0.4107$.

Step5: Calculate probability for part (d)

• There are $C(4,2)$ ways to choose 2 aces out of 4 aces and $C(4,2)$ ways to choose 2 tens out of 4 tens. $C(4,2)=\frac{4!}{2!(4 - 2)!}=6$. The number of hands with 2 aces and 2 tens is $C(4,2)\times C(4,2)=6\times6 = 36$.
• The probability $P(2\text{ aces and }2\text{ 10s})=\frac{36}{58905}\approx0.0006$.

Step6: Calculate probability for part (e)

• There are $C(4,2)$ ways to choose 2 aces out of 4 aces. There are 9 cards of the given suit, and we need to choose 2 of them. But we need to subtract the cases where the aces are already counted in the suit. The number of non - ace cards of the suit is 8. $C(4,2)=\frac{4!}{2!(4 - 2)!}=6$ and $C(8,2)=\frac{8!}{2!(8 - 2)!}=28$. The number of hands with 2 aces and 2 of the suit is $6\times28 = 168$.
• The probability $P(2\text{ aces and }2\text{ of suit})\frac{168}{58905}\approx0.0029$.

Answer:

Part (a): 0.0196
Part (b): 0.0086
Part (c): 0.4107
Part (d): 0.0006
Part (e): 0.0029