Chapter 12.7, Problem 56SR

To determine

To find: Probability that the third marble is green

Answer to Problem 56SR

Required probability $=0.04$

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 two are red and blue and not 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 two are red and blue and not replaced .

As out of available $9$ red marbles we need to pick one

So, $n\left(E\right)=9$

So, the probability that first marble to be red is $\frac{9}{29}$

Now, After first marble taken from the bag there will be only $28$ marbles.

So, $n\left(S\right)=28$

And number of red marbles remaining in bag $=8$

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 $\frac{12}{28}$

After one red and one blue marbles taken from the bag

There will be $27$ marbles in bag.

So, $n\left(S\right)=27$

And number of blue marbles remaining in bag $=11$

Given we need to pick the third marble and that needs to be green

Number of green marbles in bag $=8$

Therefore,

Probability for the third marble to be green is $\frac{8}{27}$

Therefore,

Total probability for the third marble to be red given first two are red and not replaced is

  $\begin{array}{l}P=\frac{9}{29}\times \frac{12}{28}\times \frac{8}{27}\\ \Rightarrow P=0.04\end{array}$

Therefore,

Required probability $=0.04$