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Please solve RD Sharma class 12 chapter Probability exercise 30.6 question 13 maths textbook solution

Answers (1)

Answer:

\frac{17}{400} 

Hint:

 To solve this we find the (B1, B2 and B3) then use total probability

Given:

 E1: 50% Production, E2: 25% Production, E3: 25% Production

Solution:

B1 is the event when E1 produce bulbs

B2 is the event when E2 produce bulbs

B3 is the event when E3 produce bulbs

E event when a defective bulbs is selected

\begin{aligned} &P(B_1)=\frac{1}{2} \qquad ; \qquad P(B_2)=\frac{1}{4} \qquad ; \qquad P(B_3)=\frac{1}{4}\\ &P\left (\frac{E}{B_1} \right )=\frac{4}{100}=\frac{1}{25}\\ &P\left (\frac{E}{B_2} \right )=\frac{4}{100}=\frac{1}{25}\\ &P\left (\frac{E}{B_3} \right )=\frac{5}{100}=\frac{1}{20} \end{aligned}

Using total probability theorem,

\begin{aligned} &P(E)=P(B_1)\times P\left (\frac{E}{B_1} \right )+ P(B_2)\times P\left (\frac{E}{B_2} \right )+P(B_3)\times P\left (\frac{E}{B_3} \right )\\ &P(E)=\frac{1}{2}\times \frac{1}{25}+\frac{1}{4}\times \frac{1}{25}+\frac{1}{4}\times \frac{1}{20}\\ &P(E)=\frac{17}{400} \end{aligned}

Posted by

Gurleen Kaur

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