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  • Please solve RD Sharma class 12 chapter Probability exercise 30.6 question 5 maths textbook solution

Please solve RD Sharma class 12 chapter Probability exercise 30.6 question 5 maths textbook solution

Answers (1)

Answer:

 \frac{118}{495}

Hint:

 To solve this we use total probability formula.

Given:

Bag I: 1 White,2 black and 3 red balls

Bag II: 2 White,1 black and 1 red balls

Bag III: 4 White,5 black and 3 red balls

Solution:

E1=Selecting Bag 1
E2=Selecting Bag 2
E3=Selecting Bag 3
A = Drawing a white and a red ball

As one of the bag is selected randomly,

\begin{aligned} &P(E_1)= P(E_2)=P(E_3)=\frac{1}{3}\\ &P\left ( \frac{A}{E_1} \right )=\frac{C_1\times ^3C_1}{^6C_2}=\frac{3}{15}\\ &P\left ( \frac{A}{E_2} \right )=\frac{^2C_1\times ^1C_1}{^4C_2}=\frac{2}{6}\\ &P\left ( \frac{A}{E_3} \right )=\frac{^4C_1\times ^3C_1}{^{12}C_2}=\frac{12}{66} \end{aligned}

Using total probabilty theorem

\begin{aligned} &\text { Required probability }=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{1}{3}\times \frac{3}{15}+\frac{1}{3}\times \frac{2}{6}+\frac{1}{3}\times \frac{12}{66}\\ &=\frac{1}{15}+\frac{1}{9}+\frac{2}{33}\\ &=\frac{33+55+30}{495}=\frac{118}{495} \end{aligned}

Posted by

Gurleen Kaur

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