Question

Four balls are drawn in succession without replacement from an urn containing 4 red balls, 3 blue balls and 3 yellow balls. Let Z be the random variable representing the number of blue balls.

Answer 1

$$4+3+3=10$$

There are $\dbinom{10}{4}=210$ possible outcomes.

The possible values of $Z$ are $0,1,2,3.$

$$P(Z=0)=\dfrac{\dbinom{3}{0}\dbinom{10-3}{4-0}}{\dbinom{10}{4}}=\dfrac{1(35)}{210}=\dfrac{1}{6}$$

$$P(Z=1)=\dfrac{\dbinom{3}{1}\dbinom{10-3}{4-1}}{\dbinom{10}{4}}=\dfrac{3(35)}{210}=\dfrac{1}{2}$$

$$P(Z=2)=\dfrac{\dbinom{3}{2}\dbinom{10-3}{4-2}}{\dbinom{10}{4}}=\dfrac{3(21)}{210}=\dfrac{3}{10}$$

$$P(Z=3)=\dfrac{\dbinom{3}{3}\dbinom{10-3}{4-3}}{\dbinom{10}{4}}=\dfrac{1(7)}{210}=\dfrac{1}{30}$$ $$\def\arraystretch{1.5} \begin{array}{c:c} z & 0 & 1 & 2 & 3 \\ \hline p(z) & 1/6 & 1/2 & 3/10 & 1/30 \end{array}$$