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Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.2 question 2

Answers (1)

Answer:

 \frac{1}{270725}

Hint:

\text { Probability }=\frac{\text { No. of favourable outcomes }}{ \text { Total no. of outcomes }}

Solution:

Let, A = An ace in the first draw.

        B = An ace in the 2nd draw.

        C = An ace in the 3rd draw.

        D =  An ace in the 4th draw.

\begin{aligned} &\text { Probability of A }=\frac{4}{25}\\ &=\frac{1}{13}\\ &P\left ( \frac{B}{A} \right )=\frac{3}{51}=\frac{1}{17}\\ &P\left ( \frac{C}{A\cap B} \right )=\frac{2}{50}=\frac{1}{25}\\ &P\left ( \frac{D}{A\cap B\cap C} \right )=\frac{1}{49} \end{aligned}

\begin{aligned} &\text { Required Probability }=P\left ( A\cap B\cap C\cap D \right )\\ &=P(A)\times P\left ( \frac{B}{A} \right )\times P\left ( \frac{C}{A\cap B} \right )\times P\left ( \frac{D}{A\cap B\cap C} \right )\\ &=\frac{1}{13}\times \frac{1}{17}\times \frac{1}{25}\times \frac{1}{49}\\ &=\frac{1}{270725} \end{aligned}

Posted by

Gurleen Kaur

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