We randomly draw eight cards from a deck of 52 cards. Given that three cards are spades, what is the probability that the other five are also spades?

Question

In this question, for P(B), where B is the event that at least 3 spades are drawn, why is P(B) equal to the image below?

Why isn't P(B) = (13 C 3)(49 C 5) / (52 C 8)? In my answer for P(B), I do not understand why this combination does not cover all scenarios where there are at least 3 spades and the other 5 can be anything else--spades, kings, queens, or hearts. Shouldn't (49 C 5) cover the other combinations that the P(B) sums up in the image above does?

The answer to the question, where P(A|B) = the probability, given that the first 3 cards drawn are spades, the next 5 are also spades.

Answer 1

In this question, for P(B), where B is the event that at least 3 spades are drawn, why is P(B) equal to the image below?

The event is not that a particular three from the eight cards are spades.   So you cannot say that the other five can be anything else; since you don't know which cards are "the other".

The event is merely that at least three from the eight cards are spades.  

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Consider this: Take two fair coins and flip them, the equally probable results are $\rm\{HH,HT,TH,TT\}$. The event of "the first is a head" is $\rm\{HH,HT\}$. The event of "at least one is a head" is $\rm\{HH,HT,TH\}$.

Answer 2

The problem with ${13\choose3}{49\choose5}$ is that some possibilities are counted multiple times.

For example, the hand with $2$, $3$, $4$, and $5$ of spade with 4 non-spades is counted four times.

• $2$, $3$, $4$ of spade, then $5$ of spade and 4 other;

• $2$, $3$, $5$ of spade, then $4$ of spade and 4 other;

• $2$, $4$, $5$ of spade, then $3$ of spade and 4 other;

• $3$, $4$, $5$ of spade, then $2$ of spade and 4 other.

It will be hard to modify this value to get the right one.

This is why they made the sum, conditionning on the number of spade drawn.

Answer 3

Answer : c Explanation : The Random Experiment follows hypergeometric distribution with, N = 52 since there are 52 cards in a deck.k = 13 since there are 13 spades in a deck.n = 8 since we randomly select 8 cards from the deck.x = 0 to 2 since we want less than 3 spades.

h (x < 3; N, n, k) = [kCx][N [Minus] kCn [Minus] x]/[NCn]

where x ranges from 0 to 2.

h (x < 3; 52, 8, 13) = [13 C0][39 C8]/[52 C8] + [13 C1][39 C7]/[52 C8] + [13 C2][ 39 C6]/[52 C2]

Now enter this into a CAS capable of calculating the values of the hypergeometric distribution in the notation the responsibles of your CAS did select:

h (x < 3; 52, 8, 13) = 0.685.