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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 14 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 14 maths textbook solution

Answers (1)

Answer:

X 0 1 2 3 4
P\left ( X \right ) \frac{969}{2530} \frac{455}{969} \frac{38}{253} \frac{2}{253} \frac{1}{2530}


Given: defective bolts are accidently mixed with 20 good ones. If 4 bolts are drawn at random, find the probability of defective bolts drawn.Hint: we combination formula.

\therefore X can take values 0,1,2, 3 or 4

\therefore there are total 25 bolts ( 20 good + 5 defective) bolts

N(s) = total possible ways of selecting 5 bolts = 25c_{4}

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

=\frac{20 c_{4}}{25 c_{4}}=\frac{20 \times 19 \times 18 \times 17}{25 \times 24 \times 23 \times 22}=\frac{969}{2530}  

P (x=1) =P (selecting 1 defective bolt and 3 good bolts)

=\frac{5 c_{1} \times 20 c_{3}}{25 c_{4}}=\frac{114}{969}

P (x=2)= P (selecting 2 defective bolts and 2 good bolt)

=\frac{5 c_{2} \times 20 c_{2}}{25 c_{4}}=\frac{38}{253}

P (x=3)= P (selecting 3 defective bolts and 1 good bolt)

=\frac{5 c_{3} \times 20 c_{1}}{25 c_{4}}=\frac{4}{253}

P (x=4) =P (selecting 4 defective bolts and no good bolt)

=\frac{5 c_{4}}{25 c_{4}}=\frac{1}{2530}

Now we have X and P(X)

Following table represents the probability distribution of x:

X 0 1 2 3 4
P\left ( X \right ) \frac{969}{2530} \frac{455}{969} \frac{38}{253} \frac{2}{253} \frac{1}{2530}

 

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