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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 11

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 11

Answers (1)

Answer:

 i)\: \frac{3}{19}, \quad (ii)\frac{6}{19}, \quad (iii)\frac{10}{19}

Hint:

 Use baye’s theorem.

Given:

 An insurance company is used 3000, 4000, 5000 trucks. The probability of accident involving a scooter, a car, a truck 0.02, 0.03, 0.04, one insured vehicle meets with accident. Find the probability of (i) Scooter, (ii) car, (iii)Truck

Solution:

Let E1,E2,E3 denote the event that vehicle is (i) Scooter, (ii) car, (iii) Truck.

Let A be event that vehicle meet an accident.

It is given that 3000 scooter, 4000 car and 5000 truck.

Total vehicle = 3000 + 4000 + 5000 = 12000

\begin{aligned} &P(E_1)=\frac{3000}{12000}=\frac{1}{4}\\ &P(E_2)=\frac{4000}{12000}=\frac{1}{3}\\ &P(E_3)=\frac{5000}{12000}=\frac{5}{12}\\ &P\left ( \frac{A}{E_1} \right )=0.02=\frac{2}{100}\\ &P\left ( \frac{A}{E_2} \right )=0.03=\frac{3}{100}\\ &P\left ( \frac{A}{E_3} \right )=0.04=\frac{4}{100}\\ \end{aligned}

Using Baye’s theorem we get

Required probability

\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{4}\times \frac{2}{100} }}{\frac{1}{4}\times \frac{2}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{5}{12}\times \frac{4}{100}}\\ &=\frac{\frac{1}{2}}{\frac{1}{2}+1+\frac{5}{3}}\\ &=\frac{\frac{1}{2}}{\frac{3+6+10}{6}}\\ &=\frac{3}{19} \end{aligned}

Required probability

\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}\times \frac{3}{100} }}{\frac{1}{4}\times \frac{2}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{5}{12}\times \frac{4}{100}}\\ &=\frac{1}{\frac{1}{2}+1+\frac{5}{3}}\\ &=\frac{1}{\frac{3+6+10}{6}}\\ &=\frac{6}{19} \end{aligned}

Required probability

\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{5}{12}\times \frac{4}{100} }}{\frac{1}{4}\times \frac{2}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{5}{12}\times \frac{4}{100}}\\ &=\frac{\frac{5}{3}}{\frac{1}{2}+1+\frac{5}{3}}\\ &=\frac{\frac{5}{3}}{\frac{3+6+10}{6}}\\ &=\frac{10}{19} \end{aligned}

Posted by

Gurleen Kaur

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