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Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

54. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

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alok kumar singh | Contributor-Level 10

4 months ago

54. It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement.

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

$∴ P (X = 0) = P$ (4 non-defective and 0 defective) $=^{4} C_{0} . \frac{4}{5} . \frac{4}{5} . \frac{4}{5} . \frac{4}{5} = \frac{256}{625}$

$P (X = 1) = P$ (3 non-defective and 1 defective) $=^{4} C_{1} . (\frac{1}{5}) . {(\frac{4}{5})}^{3} = \frac{256}{625}$

$P (X = 2) = P$ (2 non-defective and 2 defective) $=^{4} C_{2} . {(\frac{1}{5})}^{2} . {(\frac{4}{5})}^{2} = \frac{96}{625}$

$P (X = 3) = P$ (1 non-defective and 3 defective) $=^{4} C_{3} . {(\frac{1}{5})}^{3} . (\frac{4}{5}) = \frac{16}{625}$

$P (X = 4) = P$ (0 non-defective and 4 defective) $=^{4} C_{4} . {(\frac{1}{5})}^{4} . {(\frac{4}{5})}^{0} = \frac{1}{625}$

Therefore, the required probability distribution is as follows.

X	0	1	2	3	4
P (X)	256/625	256/625	96/625	16/625	1/625

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54. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

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