Question

Suppose three balls are drawn (with replacement) from a bag that contains 2 white balls and 3 orange balls. The number of orange balls is counted. Complete parts (a) and (b) below. (a) Determine the histogram for the number of orange balls in the sample. Choose the correct histogram below. A OB OC. A P() 1- AP(x) 1- AP(x) 1- OD. AP(x) 1- 0.5- 0.5- 0.5- 0.5- X 0 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 (b) What is the expected number of orange balls in the sample? @D (Round to two decimal places as needed.)

Answer 1

Total number of orange balls = n = 3

Probability of drawing an orange ball =p = 3/5

Let X denotes the number of orange balls.

X ~ Binomial (n=3, p=3/5 = 0.6)

$\small P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$

Therefore, putting value of x=0,1,2,3

$\small P(X=0)=\binom{3}{0}0.6^0(1-0.6)^{3-0}=0.064 \\ P(X=1)=\binom{3}{1}0.6^1(1-0.6)^{3-1}=0.288 \\ P(X=2)=\binom{3}{2}0.6^2(1-0.6)^{3-2}=0.432 \\ P(X=0)=\binom{3}{3}0.6^3(1-0.6)^{3-3}=0.216$

So, the probability distribution table is given as -

Number of orange balls (x)	Probability P(X=x)
0	0.064
1	0.288
2	0.432
3	0.216

(a) From the probability distribution table we can see that option A is correct.

(b) Expected number of orange balls :

$\small E(X)=\sum xP(X=x)$

            $\small =[0\times 0.064]+[1\times 0.288]+[2\times 0.432]+[3 \times 0.216]$

             $\small \textbf{=1.8 \ \ \ \ \ \ \ \ \ (Answer)}$