Answer to Question #154652 in Statistics and Probability for fatih

Question

A box contains ten light bulbs of which four are defective. A bulb is selected from the box and tested. If it is defective, another bulb is selected and tested, until a non-defective bulb is chosen. Let the random variable X be the number of bulbs chosen. a) Write sample space for this random experiment? b) Find probability mass Function of X? c) Find CDF of X?

Answer 1

a) Sample space: $S=\{1,2,3,4,5\}$

$X\geq 1$ , because we must choose at least one bulb;

We can select at most 5 bulbs, because we have only 4 defective bulbs and the last bulb chosen is non-defective.

b) If we selected $n$ defective bulbs, then the probability that the next one is defective is $\frac{\text{No. of defective bulbs left}}{\text{Total no. of bulbs left}}= \frac{4-n}{10-n}$ .

If we selected $n$ defective bulbs, then the probability that the next one is non-defective is $\frac{\text{No. of non-defective bulbs left}}{\text{Total no. of bulbs left}}= \frac{6}{10-n}$ .

Now we can find the probability mass Function of X:

$P(X=1)=\frac{6}{10}=\frac{3}{5}$

$P(X=2)=\frac{4}{10}\cdot \frac{6}{9}=\frac{4}{15}$

$P(X=3)=\frac{4}{10}\cdot \frac{3}{9}\cdot \frac{6}{8}=\frac{1}{10}$

$P(X=4)=\frac{4}{10}\cdot \frac{3}{9}\cdot \frac{2}{8}\cdot \frac{6}{7}=\frac{1}{35}$

$P(X=5)=\frac{4}{10}\cdot\frac{3}{9}\cdot \frac{2}{8}\cdot \frac{1}{7}\cdot \frac{6}{6}=\frac{1}{210}$

c) $F(1)=P(X\leq1)=\frac{3}{5}$

$F(2)=P(X\leq 2)=P(X=1)+P(X=2)=\frac{3}{5}+\frac{4}{15}=\frac{13}{15}$

$F(3)=P(X\leq 3)=P(X=1)+P(X=2)+P(X=3)=\frac{3}{5}+\frac{4}{15}+\frac{1}{10}=\frac{29}{30}$

$F(4)=P(X\leq 4)=P(X=1)+P(X=2)+P(X=3)+P(X=4)=\frac{3}{5}+\frac{4}{15}+\frac{1}{10}+\frac{1}{35}=\frac{209}{210}$

$F(5)=P(X\leq 5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=\frac{3}{5}+\frac{4}{15}+\frac{1}{10}+\frac{1}{35}+\frac{1}{210}=1$

Answers:

a) $S=\{1,2,3,4,5\}$

b) $\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} X & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & \tfrac{3}{5} & \tfrac{4}{15} & \tfrac{1}{10} & \tfrac{1}{35} & \tfrac{1}{210} \\ \end{array}$

c) $\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} X & 1 & 2 & 3 & 4 & 5 \\ \hline F(x) & \tfrac{3}{5} & \tfrac{13}{15} & \tfrac{29}{30} & \tfrac{209}{210} & 1 \\ \end{array}$