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}

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