Question

A bag contains $4$ green marbles, $3$ red marbles and $7$ blue marbles.

One marble is taken from the bag and put back after checking its color.

A second marble is then taken out.

What is the probability that the first is blue and the second red $?$

Answer 1

Step-1: Calculate the total ball in the box:

There are $14$ marbles $(4\text{green,}3\text{red}\text{and}7\text{blue)}$ in the box.

Therefore, the number of possible outcomes in the first draw are $14$.

There are $7$ blue marbles.

So the number of favorable outcomes are $7$.

Step-2: Find the probability that the first is blue and the second is red.

The probability is ratio of the number of favorable outcomes to the number of possible outcomes.

The probability that the first marble being blue is exactly:$=\frac{7}{14}$

$=\frac{1}{2}$

Now, this marble is returned to the bag. So, the total number of marbles in the bag remains $14$.

Hence, the number of possible outcomes in the second draw are $14$.

The number of red marbles is $3$.

Thus, the number of favorable outcomes are $3$.

The probability that the second marble is red$=\frac{3}{14}$

The probability that the first is blue and the second red is,

$$\begin{gathered} P\left(E\right)=\frac{1}{2}\times \frac{3}{14} \\ =\frac{3}{28} \end{gathered}$$

Therefore, the probability that the first is blue and the second red is $\frac{3}{28}$.