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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.1 question 3

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.1 question 3

Answers (1)

Answer:

 \frac{1}{15}

Hint:

Use Formula

\text { Probability }=\frac{\text { Required event }}{ \text { Total no. of event }}

Given, an event where two numbers appear on throwing two dice differently.

So, as per the question

Let,     A       = No on dice where two values are different
                    = { (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6)

                  (3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,6) }

Similarly,

B= Event where sum of the no. is 4

= (1,3), (2,2), (3,1)

Now,

For event where no. is different and also its sum is 4

\begin{aligned} &=A \cap B\\ &=\left \{ \left \{ 1,3 \right \}, \left \{ 3,1 \right \} \right \} \end{aligned}

Thus,

\begin{aligned} &\text { Probability }=\frac{n(A\cap B)}{ nA }\\ &=\frac{2}{30}\\ &=\frac{1}{15} \end{aligned}

Posted by

Gurleen Kaur

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