From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.

Solution

Total number of bulbs = 10

Number of defective bulbs = 3

Number of non-defective bulbs = 7

P ( drawing a defective bulb ), P = $\frac{3}{10}$

P ( drawing a non-defective bulb ), q = $\frac{7}{10}$

Two bulbs are drawn.

Let X denote the number of defective bulbs, then X can take values

0, 1, and 2.

P ( X = 0 ) = P ( drawing both non-defective bulb ) = ${(\frac{7}{10})}^{2}$

P ( x = 1 ) = P ( drawing one defective bulb and one non-defective bulbs )

= P ( drawing a non-defective bulb and a defective bulb ) + P ( drawing a defective bulb and a non-defective bulb )

$= (\frac{7}{10}) (\frac{3}{10}) + (\frac{3}{10}) (\frac{7}{10}) = \frac{21}{50} P (X = 2) = P (drawing both defective bulbs) = {(\frac{3}{10})}^{2}$

Required probability distribution is

X	0	1	2
P ( x )	$\frac{49}{100}$	$\frac{21}{50}$	$\frac{9}{100}$

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