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

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

Answers (1)

Answer:

 \frac{1}{15}, \frac{2}{5},\frac{8 }{15}

Hint:

Use baye’s theorem. 

Given:

 Suppose we have four boxes A, B, C, D.

Box

Red

White

Black

A

1

6

3

B

6

2

2

C

8

1

1

D

0

6

4

One of box has been selected at random and single marble drawn if the marble is red what is the probability that it drawn box A, B, C, D.

Solution:

Let

R →  event that red marble

A →  event select box A

B →  event select box B

C →  event select box C

D →  event select box D

\begin{aligned} &P(A)=\text { probability of box A }=\frac{1}{4}\\ &P(B)=\text { probability of box B }=\frac{1}{4}\\ &P(C)=\text { probability of box C }=\frac{1}{4}\\ &P(D)=\text { probability of box D }=\frac{1}{4}\\ &P\left ( \frac{R}{A} \right )=\frac{1}{10}\\ &P\left ( \frac{R}{B} \right )=\frac{6}{10}\\ &P\left ( \frac{R}{C} \right )=\frac{8}{10}\\ &P\left ( \frac{R}{D} \right )=\frac{0}{10}\\ \end{aligned}

Using Baye’s theorem

\begin{aligned} &P\left (\frac{A}{R} \right )=\frac{P(A).P\left ( \frac{R}{A} \right )}{P(A)\times P\left ( \frac{R}{A} \right )+P(B)\times P\left ( \frac{R}{B} \right )+P(C)\times P\left ( \frac{R}{C} \right )+P(D)\times P\left ( \frac{R}{D} \right )}\\ &=\frac{\frac{1}{4}\times \frac{1}{10} }{\frac{1}{4}\times \frac{1}{10}+\frac{1}{4}\times \frac{6}{10}+\frac{1}{4}\times \frac{8}{10}+\frac{1}{4}\times \frac{0}{10}}\\ &=\frac{\frac{1}{4}\times \frac{1}{10} }{\frac{1}{4}\times \frac{1}{10}(6+1+8)}\\ &=\frac{1}{15} \end{aligned}

Using Baye’s theorem

\begin{aligned} &P\left (\frac{B}{R} \right )=\frac{P(B).P\left ( \frac{R}{B} \right )}{P(A)\times P\left ( \frac{R}{A} \right )+P(B)\times P\left ( \frac{R}{B} \right )+P(C)\times P\left ( \frac{R}{C} \right )+P(D)\times P\left ( \frac{R}{D} \right )}\\ &=\frac{\frac{1}{4}\times \frac{6}{10} }{\frac{1}{4}\times \frac{1}{10}+\frac{1}{4}\times \frac{6}{10}+\frac{1}{4}\times \frac{8}{10}+\frac{1}{4}\times \frac{0}{10}}\\ &=\frac{\frac{1}{4}\times \frac{6}{10} }{\frac{1}{4}\times \frac{1}{10}(6+1+8)}\\ &=\frac{16}{15}\\ &=\frac{2}{5} \end{aligned}

Using Baye’s theorem

\begin{aligned} &P\left (\frac{C}{R} \right )=\frac{P(C).P\left ( \frac{R}{C} \right )}{P(A)\times P\left ( \frac{R}{A} \right )+P(B)\times P\left ( \frac{R}{B} \right )+P(C)\times P\left ( \frac{R}{C} \right )+P(D)\times P\left ( \frac{R}{D} \right )}\\ &=\frac{\frac{1}{4}\times \frac{8}{10} }{\frac{1}{4}\times \frac{1}{10}+\frac{1}{4}\times \frac{6}{10}+\frac{1}{4}\times \frac{8}{10}+\frac{1}{4}\times \frac{0}{10}}\\ &=\frac{\frac{1}{4}\times \frac{8}{10} }{\frac{1}{4}\times \frac{1}{10}(6+1+8)}\\ &=\frac{8}{15} \end{aligned}

Posted by

Gurleen Kaur

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