Question

Exercise 3.8.18: How many ways are there to choose five cards from 52 so that you have all the aces? How many ways are there to choose five cards from 52 so that you have four cards with the same number?

Exercise 3.8.20: How many ways are there to choose five cards from 52 so that you have two cards with the same number? Hint: This problem is more tricky than you think.

Answer 1

Total cards = 52

total aces = 4

total number of non-aces = 52

P(choosing five cards from 52 so that you have all the aces) = 48C1 * 4C4 / 52C5= 1/54145

P(chosing five cards from 52 so that you have four cards with the same number) = 13 * P(choosing five cards from 52 so that you have all the aces) = 13 * 1/54145 = 1/4165

Explanation - Similar to ace , every other 12 numbers will have same probability to be chosen as 4 same cards and one different

Exercise 3.8.20

P(chosing a particular numbered card ) =13/52

P(chosing same card as the previous one ) = 12/51

P(chosing five cards from 52 so that you have two cards with the same number) = {13 * 4*3*48*/44*40 } / {52*51*50*49*48} = 176/4165

Explanation = denominator is calculated normally as there would be 52C5 number of ways to chose 5 cards from 52 total cards . Talking about numberator ,  there are 13 different numbers from which 2 similar numbered cards can be chosen and rest different number cards are chosen from remaining 12 different numbered cards  , so after 2 similar cards , remaining would be a choice from 48 cards this is 12*4 and then 44 cards that is 11*4 and then the choice will be from 40 cards that is 10*4 . we have done this as the question states that only 2 cards will be same numbered.