Question

A bag contains 4 red and 6 black balls. A sample of 4 balls is selected from the bag without replacement. Let X is the number of red balls. Find the p.d for X? a) Find probability of X greater than 2? b) Find probability of X for non-negative integers? 

Answer 1

Let $X=$ the number of red balls.

$$P(X=0)=\dfrac{\dbinom{4}{0}\dbinom{6}{4}}{\dbinom{10}{4}}=\dfrac{1(15)}{210}=\dfrac{1}{14}$$

$$P(X=1)=\dfrac{\dbinom{4}{1}\dbinom{6}{3}}{\dbinom{10}{4}}=\dfrac{4(20)}{210}=\dfrac{8}{21}$$

$$P(X=2)=\dfrac{\dbinom{4}{2}\dbinom{6}{2}}{\dbinom{10}{4}}=\dfrac{6(15)}{210}=\dfrac{3}{7}$$

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

$$P(X=4)=\dfrac{\dbinom{4}{4}\dbinom{6}{0}}{\dbinom{10}{4}}=\dfrac{1(1)}{210}=\dfrac{1}{210}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 & 4 \\ \hline p(x) & 1/14 & 8/21 & 3/7 & 4/35 & 1/210 \\ \end{array}$$

a)

$$P(X>2)=P(X=3)+P(X=4)$$

$$=\dfrac{4}{35}+\dfrac{1}{210}=\dfrac{5}{42}$$

b)

$$P(X=0)=\dfrac{1}{14}$$

$$P(X=1)=\dfrac{8}{21}$$

$$P(X=2)=\dfrac{3}{7}$$

$$P(X=3)=\dfrac{4}{35}$$

$$P(X=4)=\dfrac{1}{210}$$