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  • Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 30 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 30 maths textbook solution

Answers (1)

Answer: k=\frac{1}{8}   \left ( i \right )\frac{1}{8}    \left ( ii \right )\frac{5}{8}    \left ( iii \right )\frac{7}{8}

Hint: Let P(X=0) = 0

P(X=x)=\left\{\begin{aligned} k x, & \text { if } x=0,1 \\ 2 k x, & \text { if } x=2 \\ k(5-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { if } x>4 \end{aligned}\right.

            Where k is a positive constant

find the value of k. also find the probability that you will get admission in

\left ( i \right ) Exactly on colleges         \left ( ii \right ) at most 2 colleges     \left ( iii \right )at least 2 colleges

Solution:

P(X=x)=\left\{\begin{aligned} k x, & \text { if } x=0,1 \\ 2 k x, & \text { if } x=2 \\ k(5-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { if } x>4 \end{aligned}\right.

Our variable is X and from equation we see that  it is taking values X = 0,1,2,3,4….( any whole number )

And it represents the no. of colleges in which you are going to get admission

According to equation given we have:

\begin{aligned} &\therefore P(x=0)=k \times 0=0 \\ &P(x=1)=k \times 1=k \\ &P(x=2)=2 k \times 2=4 k \\ &P(x=3)=k(5-3)=2 k \\ &P(x=4)=k(5-4)=k \\ &P(x>4)=0 \end{aligned}

As in question it is not given either x is random variable or the given distribution is a probability distribution, we can’t apply anything directly.

But we have a hint in the question

As P(x=0)=0

It means the chance of not getting admission in any college 0

∴admission is sure

Hence the sum of all probability must be equal to as getting admission has become a sure event.

This makes the given distribution a probability distribution an X a random variable also.

\begin{aligned} &\therefore k+4 k+2 k+k=1 \Rightarrow 8 k=1 \\ &k=\frac{1}{8} \end{aligned}

i} P (getting admission in exactly one colleges )= P\left ( x=1 \right )

=k.1=\frac{1}{8}

ii} P (getting admission in almost 2 colleges) =P\left ( x\leq 2 \right )

\begin{aligned} &=P(x=1)+P(x=2) \\ &=k+4 k=5 k=\frac{5}{8} \end{aligned}

iii} P  (getting admission in at least 2 colleges) =P\left ( x\geq 2 \right )

\begin{aligned} &=P(x=2)+P(x=3)+P(x=4) \\ &=4 k+2 k+k=7 k=\frac{7}{8} \end{aligned}

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