Answer to Question #156624 in Statistics and Probability for Mia Bianca

Question

Let D representing the number of defective units when 5 units are chosen at random from a batch of ten flash drives which contains six defective items. Find the value of random variable D.

Answer 1

Let $D$ be a random var iable defined as ,

$D(x)=$ Number of defective flash drives

Then $D$ can take the values $0,1,2,3,4,5$ .

Now corresponding to each values of $D$ we get their probability such as,

$P(D=0)=0$

$P(D=1)=\frac{^6C_1×^4C_4}{^{10}C_5}=\frac{6×1}{252}=\frac{1}{42}$

$P(D=2)=\frac{^6C_2×^4C_3}{^{10}C_5}=\frac{60}{252}=\frac{5}{21}$

$P(D=3)=\frac{^6C_3×^4C_2}{^{10}C_5}=\frac{120}{252}=\frac{10}{21}$

$P(D=4)=\frac{^6C_4×^4C_1}{^{10}C_5}=\frac{60}{252}=\frac{5}{21}$

$P(D=5)=\frac{^6C_5}{^{10}C_5}=\frac{6}{252}=\frac{1}{42}$

Which is required value of random variable $D$ .