Chapter 12.7, Problem 55SR

To determine

To find: Probability that the Second marble is blue

Answer to Problem 55SR

Required probability $=0.114$

Explanation of Solution

Given information:

A bag contains

  $12$ blue marbles , $9$ red marbles and $8$ green marbles.

Marbles are drawn one at a time

First marble is green and replaced .

Given

A bag contains

  $12$ blue marbles , $9$ red marbles and $8$ green marbles.

It means

Total number of marbles in bag $=12+8+9$

  $\Rightarrow$ Total number of marbles in bag $=29$

We know

Probability $=\frac{n\left(E\right)}{n\left(S\right)}$

  $n\left(E\right)=$ number of favourable outcomes

  $n\left(S\right)=$ Total number of events $=29$

Given

First marble is green and replaced .

As out of available $8$ green marbles we need to pick one

So, $n\left(E\right)=8$

So, the probability that first marble to be green is $\frac{8}{29}$

Given that again that first marble is replaced.

Now, After first marble taken and replaced from the bag there will be again $29$ marbles in bag.

So, $n\left(S\right)=29$

Number of blue marbles in bag $=12$

And out of them we need to pick one blue marble

So, $n\left(E\right)=12$

Therefore,

The proability for the second marble to be blue is $\frac{12}{29}$

Therefore,

Total probability for the second marble to be blue given first marble is green and replaced is

  $\begin{array}{l}P=\frac{12}{29}\times \frac{8}{29}\\ \Rightarrow P=0.114\end{array}$

Therefore,

Required probability $=0.114$