A bag contains one red, two blue, three green, and four yellow balls. A sample of three balls is taken without replacement. Let $Y$ be the number of yellow balls in the sample. Find the probability of $Y=0$, $Y=1$, $Y=2$ $Y=3$

Attempt 1

all three are yellow would be

$$ \frac{\binom{4}{3}}{\binom{10}{3}}=\frac{4}{120}=\frac{1}{30} $$

two are yellow combinations would be
yy red

$$ \binom{4}{2}\binom{1}{1} =6$$

yy green $$ \binom{4}{2}\binom{3}{1}=18 $$

yy blue

$$ \binom{4}{2}\binom{2}{1} = 12$$

adding them together and dividing getting $P(Y=2)=.3$

for one yellow

it would be

combinations for Yellow red blue

combinations for Yellow red green

combinations for Yellow b b

combinations for Yellow b g

combinations for Yellow g g

adding them all together and dividing by $120$

for no yellows

just keep going???Better algorithm?


I will first compute $Pr(Y=3)=\frac{(4)(3)(2)}{10(9)(8)}$ of which you have done so.

and then I will compute $Pr(Y=0)=\frac{6(5)(4)}{(10)(9)(8)}$

Then I will compute $Pr(Y=1)= 3 \times \frac{4(6)( 5)}{10(9)(8)}$

$Pr(Y=2)$ can be computed via sum of probability is equal to $1$.

Remark: The concern is yellow and non-yellow, we can group all the non-yellow together.


We have four yellow balls and $1 + 2 + 3 = 6$ balls that are not yellow. The number of ways we can select exactly $k$ yellow balls and $3 - k$ balls that are not yellow is $$\binom{4}{k}\binom{6}{3 - k}$$ Since there are $\binom{10}{3}$ possible selections of three of the ten balls, the desired probabilities are \begin{align*} P(Y = 0) & = \frac{\dbinom{4}{0}\dbinom{6}{3}}{\dbinom{10}{3}}\\ P(Y = 1) & = \frac{\dbinom{4}{1}\dbinom{6}{2}}{\dbinom{10}{3}}\\ P(Y = 2) & = \frac{\dbinom{4}{2}\dbinom{6}{1}}{\dbinom{10}{3}}\\ P(Y = 3) & = \frac{\dbinom{4}{3}\dbinom{6}{0}}{\dbinom{10}{3}} \end{align*}


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