For question 75, I've calculated the P(flush) a number of different ways and I keep getting 0.00198.  The answer key says 0.0042%
I am using 4 * 13C5/52C5

Dear Nadine Dame,

Thank you for you comment. The solution to #75 has been updated in the CW-HW file above to be 0.198% = 0.00198.

Please let us know if more information is needed or any questions arise.

Kind regards,

Audra

For question 86, I am getting 51.9%  Of course, I am calculating the probability of getting just one pair, if you are dealt 5 cards.  

I thought: 13 * 4C2 * 48C3 / 52C5. 

How do you get 59.91% ?

Dear Nadine Dame,

Thank you for your comment. The solution to #86 has also been updated. However, the correct answer is 0.4225 or 0.423 = 42.25% or 42.3%. It is found using the following calculation:
(13 * 4C2 * 12C3 * 4C1 * 4C1 * 4C1) / 52C5
The reasoning behind it is as follows:
- The 13 out in front represents the 13 different ranks of cards (e.g., king queen, jack 10, 9, 8, 7, 6, 5, 4, 3, 2, Ace) for which one could get one pair.
- The 4C2 represents the pair.
- The 12C3 represents the remaining 12 different ranks remaining to choose from, and we are choosing 3 of them.
- The remaining 4C1 values represent the number of ways one can get each remaining rank card. 

The original 48C3 does not rule out the possibility of getting another pair, or say a three of a kind (resulting in a full house), whereas, the more detailed explanation provided above rules out these extra hands.

Please let us know if more information is needed or any questions arise.

Kind regards,

Audra