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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 15

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 15

Answers (1)

Answer:

 \frac{20}{43}

Hint:

 Use Baye’s theorem.

Given:

 There are three coins. One is two headed coin another is biased coin that come up head 75% of time and third is also a biased coin that come up tall 40% of time. One of three coins is chosen at random and tossed and it show head. What is the probability that it was two head coins.

Solution:

 E1 be event of selecting of two headed coin

E2  be event selecting biased coin that comes up head 75% of time.

E3  be event selected biased coin that come up 40% of time tail.  

A be the event that head comes.

Then,

\begin{aligned} &P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}\\ &P\left ( \frac{A}{E_1} \right )=\text { probability of getting head on coin is two headed }=1\\ &P\left ( \frac{A}{E_2} \right )=\text { probability of getting biased coin that come 75% of head }\\ &=\frac{75}{100}=\frac{3}{4}\\ &P\left ( \frac{A}{E_3} \right )=\text { probability of getting biased coin that come 40% of head }\\ &=\frac{40}{100}=\frac{2}{5}\\ \end{aligned}

Using Baye’s theorem we get

\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}{3}\times 1 }}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{2}{5}}\\ &=\frac{20}{43} \end{aligned}

Posted by

Gurleen Kaur

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