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?
Best Answer
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.