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Need solution for RD Sharma maths class 12 chapter Probability exercise 30.2 question 13

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Answer:

 \begin{aligned} &\frac{44}{91} \end{aligned}

Hint:

\begin{aligned} &\text { Probability }=\frac{\text { No. of event favourable outcomes }}{ \text { Total no. of outcomes }} \end{aligned}

Given,

            A=A good orange in the 1st draw

            B=A good orange in the 2nd draw

            C=A good orange in the 3rd draw

Now,

\begin{aligned} &P(A)=\frac{12}{15}=\frac{4}{5}\\ &P\left ( \frac{B}{A} \right )=\frac{11}{14}\\ &P\left ( \frac{C}{B\cap A} \right )=\frac{10}{13}\\ &\text { Required Probability }=P(A\cap B\cap C)\\ &=P(A)\times P\left ( \frac{B}{A} \right )\times P\left ( \frac{C}{B\cap A} \right )\\ &=\frac{4}{5}\times \frac{11}{14}\times \frac{10}{13}\\ &=\frac{440}{910}=\frac{44}{91} \end{aligned}

Posted by

Gurleen Kaur

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