Two balls are drawn at random one by one with replacement

by

Question:-

Two balls are drawn at random one by one with replacement from an urn containing equal number of red balls and green balls. Find the probability distribution of number of red balls. Also, find the mean of the random variable.

Solution:-

Let X represent the number of red balls

Possible value of X = 0, 1, 2

Probability of red ball P(R) = \frac{1}{2}

Probability of green ball P(G) =\frac{1}{2}

P(X=0)=P(G)\times P(G)
\quad \Rightarrow \frac{1}{2}\times \frac{1}{2}= \frac{1}{4}

P(X=1)=P(G)\times P(R)+P(R)\times P(G)

\Rightarrow \quad \frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}

\Rightarrow \frac{1}{4}+\frac{1}{4}=\frac{2}{4}

\Rightarrow \frac{1}{2}

P(X=2)=P(R)\times P(R)

\Rightarrow \quad \frac{1}{2}\times \frac{1}{2}

\Rightarrow \quad \frac{1}{4}

\begin{array}{|c|c|c|c|} \hline { X } & { 0 } & { 1} & 2\\ \hline P(X) & \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\ \hline \end{array}

Mean = \Sigma X.P(X)=0\times \frac{1}{4}+1\times \frac{1}{2}+2\times\frac{1}{4}

=0+\frac{1}{2}+\frac{1}{2}

=1

Question:-A and B throw a die alternatively till one of them gets ‘6’ and win the game. Find their respective probabilities of winning, If A starts first

Solution:- For solution click here

 

Case study problem probability 5 chapter 13 class 12
Mahindra tractors is India’s leading farm equipment manufacturer. It is the largest tractor selling factory

gmath.in

Leave a Comment