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A six-card hand is dealt from an ordinary deck of cards. Find the probability that:

  1. All six cards are hearts.

With combinations, this would be all the ways you can choose 6 cards from 13, divided by all the ways you can choose 6 cards from 52:

choose(13,6)/choose(52,6)
## [1] 8.428903e-05

\[ P(six\ hearts)=\frac{\binom{13}{6}}{\binom{52}{6}}\\ P(six\ hearts)=\frac{13!}{6!7!}\cdot \frac{6!46!}{52!}\\ P(six\ hearts)=\frac{(13)(12)(11)(10)(9)(8)}{(52)(51)(50)(49)(48)(47)}\\ P(six\ hearts)= 0.00008428903 \]

  1. There are three aces, two kings, and one queen.

The total number of combinations fitting this criteria would be all the ways you can choose 3 out of 4 aces, times all the ways you can choose 2 out of 4 kings, times all the ways you can choose 1 queen out of 4.

choose(4,3)*choose(4,2)*choose(4,1)/choose(52,6)
## [1] 4.71547e-06

With combination notation: \[ Probability = \frac{\binom{4}{3}\binom{4}{2}\binom{4}{1}}{\binom{52}{6}} \\ =\frac{4!}{3!1!}\cdot \frac{4!}{2!2!}\cdot \frac{4!}{1!3!}\cdot \frac{6!46!}{52!}\\ =\frac{(4)(6)(4)(6!)}{(52)(51)(50)(49)(48)(47)}\\ =0.00000471547 \]

  1. There are three cards of one suit and three of another suit

The total number of combinations fitting this criteria would be all the ways you can choose two out of four suits, times all the ways you can choose three cards from one suit, times all the ways you can choose three cards from another.

choose(4,2)*choose(13,3)*choose(13,3)/choose(52,6)
## [1] 0.02410666

\[ Probability = \frac{\binom{4}{2}\binom{13}{3}\binom{13}{3}}{\binom{52}{6}} \\ =\frac{4!}{2!2!}\cdot \frac{13!}{3!10!}\cdot \frac{13!}{3!10!}\cdot \frac{6!46!}{52!}\\ =\frac{(13)(12)(11)(13)(12)(11)(6)(5)(4)}{(52)(51)(50)(49)(48)(47)}\\ =0.02410666 \]