A box contains 12 items of which 3 are defective A sample of 3 items is selected from the box - Maths - Probability

Question

A box contains 12 items of which 3 are defective A sample
of 3 items is selected from the box Let X denotes the number of
defective items in a sample Find the probability distribution of
X

Ajanta Trivedi · Accepted Answer

total number of items = 12

defective items = 3 , remaining items = 12-3 =9

since the sample of 3 items are selected from the box.

let X denote the random variable number of defective items in the sample.

therefore X can take values 0 , 1, 2 , 3.

P( X = 0) = P(getting no defective item) = $\frac{{}^{9}{C_{3}}}{{}^{12}{C_{3}}} = \frac{9!}{3! \cdot 6!} * \frac{3! \cdot 9!}{12!} = \frac{9 * 8 * 7}{12 * 11 * 10} = \frac{21}{55}$

P( X =1) = P(getting 1 defective and 2 non defective item) = $\frac{{}^{3}{C_{1}} \cdot {}^{9}{C_{2}}}{{}^{12}{C_{3}}} = \frac{3 * 9 * 8}{2} * \frac{3 * 2}{12 * 11 * 10} = \frac{27}{55}$

P( X =2) = P(getting 2 defective and 1 non defective item) = $\frac{{}^{3}{C_{2}} \cdot {}^{9}{C_{1}}}{{}^{12}{C_{3}}} = \frac{3 * 9}{1} * \frac{3 * 2}{12 * 11 * 10} = \frac{27}{220}$

P( X =3) = P(getting all three defective items ) = $\frac{{}^{3}{C_{3}}}{{}^{12}{C_{3}}} = \frac{1 * 3 * 2}{12 * 11 * 10} = \frac{1}{220}$

therefore the probability distribution for random variable X is:

X	0	1	2	3
P(X)	$\frac{21}{55}$	$\frac{27}{55}$	$\frac{27}{220}$	$\frac{1}{220}$

hope this helps you.

cheers!!

A box contains 12 items of which 3 are defective.A sample of 3 items is selected from the box.Let X denotes the number of defective items in a sample.Find the probability distribution of X.