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Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 27 maths textbook solution

Answers (1)

Answer:                

X 0 1 2
P\left ( X \right ) \frac{7}{15} \frac{7}{15} \frac{1}{15}


Hint: Use probability & combination formula

Given: 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: X represents the number of defective bulbs drawn

∴X can take values 0, 1 or 2 (as maximum bulbs drawn are 2)

∴ there are total 10 bulbs (7 good + 3 defectives)

n(s) = total possible ways of selecting ways of selecting 2 items from sample = 10c_{2}

P(x=0) = P (selecting no defective bulb)

=\frac{7 c_{2}}{10 c_{2}}=\frac{7 \times 6}{10 \times 9}=\frac{7}{15}

P(x=1) = P (selecting 1 defective bulb and 1 good bulb)

=\frac{7 c_{1} \times 3 c_{1}}{10 c_{2}}=\frac{7}{15}

P(x=2) = P (selecting 2 defective bulb and 0 good bulb)

=\frac{7 c_{0} \times 3 c_{2}}{10 c_{2}}=\frac{3 \times 2}{10 \times 9}=\frac{1}{15}

X 0 1 2
P\left ( X \right ) \frac{7}{15} \frac{7}{15} \frac{1}{15}
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