Question

Thirteen cards are dealt from a well-shuffled standard deck. What is the probability of getting: a) all red cards
b) 7 diamonds and 6 hearts
c) at least 1 face card
d) all face cards
(Please leave answers in factorial form & show work because I need to know how to do this)

Answer 1

For a, you're looking for all red cards when you deal thirteen cards out. The probability of getting a red card on the first card is \(26/52\). The second card's probability will be \(25/51\). It will continue like so until the 13th card, where the probability of getting a red card will be \(14/40\). When you multiply those all together, you get \(\dfrac{26\cdot25\cdot\ldots\cdot14}{52\cdot51\cdot\ldots\cdot40}\). To write that in factorial notation, the numerator you'll want \(26!\), but without the \(13!\) part, so you'll put the \(13!\) in the denominator. The denominator you'll want \(52!\), but without the \(39!\) part, so you'll put the \(39!\) in the numerator. So to finish up, you'll have \(\dfrac{26!\cdot39!}{52!\cdot13!}\)

Answer 2

Problem d is basically the same as a, so you should be able to work it out the same way. Problem c is a bit different. The easiest way to find the probability of getting at least one face card is to find the probability of getting no face cards and invert it. So, for the first card the probability of not getting a face card is \(40/52\), and it will continue down to \(28/40\). So the probability of no face cards is \(\dfrac{40!39!}{52!27!}\), making the probability of getting at least one face card \(1-\dfrac{40!39!}{52!27!}\).