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  • Need solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise Case Study Based  Question, question 5 sub question  (v).

Need solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise Case Study Based  Question, question 5 sub question  (v).

Answers (1)

Answer:

            (d)

Hint:

            You must know rules of finding probability

Given:

            Machine A produce 1% defective item, B produce 5%, C produce 7%. Operator A on job for 50% of time B on job for 30% and C on job for 20%

Solution:

\begin{aligned} &P(A)=0.5, P(B)=0.3, P(C)=0.2 \\ &P\left(\frac{D}{A}\right)=0.01, P\left(\frac{D}{B}\right)=0.05, P\left(\frac{D}{C}\right)=0.07 \end{aligned}

A= Event that item produced by operator A

B= Item produced by operator B

C= Item produced by operator C

D= Item produced is defective

Defective item is produced, that the probability it was produced by B or C

P\left(\frac{B}{D}\right)+P\left(\frac{C}{D}\right)

=\frac{P(B) \cdot P\left(\frac{D}{B}\right)}{P(A) \cdot P\left(\frac{D}{A}\right)+P(B) \cdot P\left(\frac{D}{B}\right)+P(C) \cdot P\left(\frac{D}{c}\right)}+\frac{P(C) \cdot P\left(\frac{D}{C}\right)}{P(A) \cdot P\left(\frac{D}{A}\right)+P(B) \cdot P\left(\frac{D}{B}\right)+P(C) \cdot P\left(\frac{D}{C}\right)}

\begin{aligned} &=\frac{0.3 \times 0.05}{0.5 \times 0.01+0.3 \times 0.05+0.2 \times 0.07}+\frac{0.2\times 0.07}{0.5 \times 0.01+0.3 \times 0.05+0.2 \times 0.07} \\ &=\frac{0.015}{0.034}+\frac{0.014}{0.034} \\ &=\frac{0.029}{0.034} \\ &=\frac{29}{34} \end{aligned}

So, (d) is correct answer

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