Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 8 maths

Answers (1)

Answer:

 i)\: \frac{12}{17}, \quad ii)\: \frac{5}{17}

Hint:

 Use Baye’s theorem.

Given:

 A letter is known to have come either from LONDON or CLIFTON on the envelope just two consecutive letter on are visible what is probability that letter has come from (i) LONDON (ii) CLIFTON

Solution:

Let E1 is event that letter come from LONDON and E2 is event that letter come from CLIFTON.

Thus

P(E_1)=P(E_2)=\frac{1}{2}

Let A be event come 2 consecutive letter on envelope are ON.

Then

P\left ( \frac{A}{E_1} \right )=\frac{2}{5}

And

P\left ( \frac{A}{E_2} \right )=\frac{1}{6}

Now

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\left (\frac{E_1}{A} \right )=\frac{{\frac{1}{2}\times \frac{2}{5} }}{\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{1}{6}}\\ &=\frac{12}{17} \end{aligned}

i. Now

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 )}\\ &P\left (\frac{E_2}{A} \right )=\frac{{\frac{1}{2}\times \frac{1}{6} }}{\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{1}{6}}\\ &=\frac{\frac{1}{6}}{\frac{12+5}{30}}\\ &=\frac{1}{6}\times \frac{30}{17}\\ &=\frac{5}{17} \end{aligned}

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