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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.6 question 11

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.6 question 11

Answers (1)

Answer:

 \frac{59}{130}

Hint:

 To solve this we use

\begin{aligned} &P(A)=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 ) \end{aligned}

Given:

 An urn contains 10 white and 3 black balls. Another urn contains 3 white and 5 black balls.

Solution:

E1=2W …{2 white balls are transferred from I to II urn}
E2=2B……{2 black balls are transferred from I to II urn}
E3=1W,1B….{A white ball and a black ball are transferred from I to II urn}
A = White ball drawn
\begin{aligned} &P(E_1)=\frac{^{10}C_2}{^{13}C_2} \qquad ; \qquad P(E_2)=\frac{^{3}C_2}{^{13}C_2} \qquad ; \qquad P(E_3)=\frac{^{10}C_1\: ^{3}C_1}{^{13}C_2}\\ &A=\text { white ball }\\ &P\left (\frac{A}{E_1} \right )=\frac{1}{2} \qquad ; \qquad P\left (\frac{A}{E_2} \right )=\frac{3}{10} \qquad ; \qquad P\left (\frac{A}{E_3} \right )=\frac{2}{5} \end{aligned}

Using total probabilty theorem

\begin{aligned} &\text { required probability }=P(A)=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{10\times 9}{13\times 12}\times \frac{1}{2}+\frac{3\times 2}{13\times 12}\times \frac{3}{10}+\frac{30\times 2}{13\times 12}\times \frac{2}{5}\\ &=\frac{15}{52}+\frac{3}{260}+\frac{2}{13}\\ &=\frac{225+9+120}{780}=\frac{354}{780}=\frac{177}{390}\\ &=\frac{59}{130} \end{aligned}

 

Posted by

Gurleen Kaur

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