Three cards are drawn in succession from a deck without replacement. Find the probability distribution for the number of spades.
Find the probability distribution for the number of spades.
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Let the random variable X be number of spades then using bionomial distribution we can calculate
P(X=r) = probablilty that r spades comes out of n cards then we can write it as nCr * p^r * q^(n-r) where n is the total number of cards drawn and r is the number of spades and p is the probability of success and q is the probability of failure
Probability that the drawn card is spade = p = Probability of success i.e. 13/52
Probability that the drawn card is not spade = q = 1-p = Probability of failure i.e. 39/52
Therefore P(X=r) = 3Cr * (13/52)^r * (39/52)^(n-r)
Substituting values of r you can get probability of getting r spades which gives us the probability distribution for the number of spades.
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Let the random variable X be number of spades then using bionomial distribution we can calculate As there are 13 spade cards out of 52 cards therefore if we choose $r$ cards form spade then we have to choose $(3-r)$ cards that are non spade.
$P(X=r)$ = probablilty that $r$ spades comes out of 3 cards then we can write it as $$\frac{(13Cr * 39C(3-r))}{52C3}$$
Substituting values of $r$ you can get probability of getting $r$ spades which gives us the probability distribution for the number of spades.
Let random variable $X$ be the number of spades. We want to calculate $\Pr(X=0)$, $\Pr(X=1)$, $\Pr(X=2)$, and $\Pr(X=3)$.
As a sample, we calculate $\Pr(X=2)$.
There are $\binom{52}{3}$ three-card hands, all equally likely. We count the number of two-spade hands.
The $2$ spades can be chosen in $\binom{13}{2}$ ways. For each of these ways, there are $\binom{39}{1}$ ways to choose the non-spade, for a total of $\binom{13}{2}\binom{39}{1}$.
Thus $\Pr(X=2)=\dfrac{\binom{13}{2}\binom{39}{1}}{\binom{52}{3}}$.
Another way: The event $X=2$ can happen in $3$ different ways: SSN (spade, then spade, then non-spade); SNS; NSS.
We calculate the probability of SSN. The probability the first card is a spade is $\frac{13}{52}$. Given that the first card was a spade, the probability the next is a spade is $\frac{12}{51}$. Given the first two cards are spades, the probability the third is a non-spade is $\frac{39}{50}$. Thus the probability of SSN is $\frac{13}{52}\cdot \frac{12}{51}\cdot \frac{39}{50}$.
Similarly, calculate the probability of SNS and NSS. (Or else note that it is clear that each is the same as the probability of SSN.) Add up.