Please solve RD Sharma class 12 chapter Probability exercise 30.7 question 5 maths textbook solution

Answers (1)

Answer:

\frac{8}{11} 

Hint:

Use baye’s theorem. 

Given:

 Suppose a girl throw a die. If she gets 1, 2 she toss a coin three times and note the number of tails. If she get 3, 4, 5, 6 she toss a coin once and note whether a head or tail is obtained. If she obtained exactly one tail, what is probability she threw 3, 4, 5, 6 with die.

Solution:

Let E1  be event that girl get 1 or 2.

Thus

P(E_1)=\frac{2}{6}=\frac{1}{3},P(E_2)=\frac{4}{6}=\frac{2}{3},P(E_3)=\frac{3}{8}

Let E2  be event that obtain tail. If she tossed a coin 3 time and exactly 1 tail show up then total favorable.

{ (THH), (HTH), (HHT) }=3

P\left ( \frac{A}{E_1} \right )=\frac{3}{8}

If she tossed the coin once and exactly 1 tail show up then total favorable.

P\left ( \frac{A}{E_2} \right )=\frac{1}{2}

The probability that she threw 3, 4, 5, 6 with given that she got exactly one tail or in other word

P\left ( \frac{A}{E_2} \right )

Using Baye’s theorem we get

\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 )}\\ &=\frac{{\frac{2}{3}\times \frac{1}{2} }}{\frac{1}{3}\times \frac{3}{8}+\frac{2}{3}\times \frac{1}{2}}\\ &=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}\\ &=\frac{8}{11} \end{aligned}

 

 

 

 

 

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