Question

WILL MARK AS BRAINLIEST 4. Suppose there is a card game where you are dealt a hand of three cards. You have already learned that the total number of three-card hands that can be dealt from a deck of 52 cards is: 52C3=52!/49!3! 52C3=22100 Calculate the probability of getting a hand that has exactly two aces in it (A A X). Do this by finding out the number of possible hands that have exactly two aces, and then dividing by the total possible number of three-card hands that is stated above. Part A: Use the multiplication principle to tell the total number of three-card hands (permutations) that can be made with two aces. (2 points) Part B: In the answer from Part I, each two-ace hand got counted twice. For example, A A X got counted as a separate hand from A A X. Since order should not matter in a card hand, these are really the same hand. What is the actual number of two-ace hands (combinations) you can get from a deck of 52 cards?(2 points) Part C: Find the probability of drawing a three-card hand that includes two aces from a deck of 52 cards. Write your answer as a fraction. (2 points)

Answer 1

Answer:

Part A- 6

Part B- 3

Part C- 3/22100

Step-by-step explanation:

Part A-

Use the permutation formula and plug in 3 for n and 2 for k.

nPr=n!/(n-k)!

3P2=3!/(3-2)!

Simplify.

3P2=3!/1!

3P2=6

Part B-

Use the combination formula and plug in 3 for n and 2 for k.

nCk=n!/k!(n-k)!

3C2=3!/2!(3-2)!

Simplify.

3C2=3!/2!(1!)

3C2=3

Part C-

It is given that the total number of three-card hands that can be dealt from a deck of 52 cards is 22100. Use the fact that the probability of something equals the total successful outcomes over the sample space. In this case the total successful outcomes is 3 and the sample space is 22100.

I believe the answer is 3/22100

I honestly suck at probability but I tried my best.