Question

Suppose you have a bag containing $3$ blue and $5$ red marbles. If you pick a marble at random and then pick another marble without replacing, find the probability of picking a blue marble both the times.

• A. $\frac{3}{8}$
  
• B. $\frac{3}{28}$
  
• C. $\frac{2}{7}$
  
• D. $\frac{25}{28}$
  

Answer 1

The correct option is B $\frac{3}{28}$


Number of blue marbles $=3$
Total number of marbles
$=3+5$
$=8$

The probability $\text{P(B1)}$ of picking up a blue marble in the first draw is:

$\text{P(B1)}=\frac{\text{Number of blue marbles}}{\text{Total number of marbles}}$

$\text{P(B1)}=\frac{3}{8}$

After picking up one blue marble, the marble is not replaced in the bag.
Now,
Number of blue marbles $=2$
Total number of marbles $=7$

The probability $\text{P(B2)}$ of picking up a blue marble in the second draw is:

$\text{P(B2)}=\frac{\text{Number of blue marbles}}{\text{Total number of marbles}}$

$\text{P(B2)}=\frac{2}{7}$

The probability $\text{P(B1 and B2)}$ of picking up a blue ball in first draw and then picking up a blue ball in the second draw is obtained by multiplying $\text{P(B1)}$ and $\text{P(B2)}$ as these two events are $\text{Dependent Events}$.

$\text{P(B1 and B2)}=\text{P(B1)}\times \text{P(B2)}$

$\text{P(B1 and B2)}=\frac{3}{8}\times \frac{2}{7}$

$\text{P(B1 and B2)}=\frac{3}{28}$


The probability $\text{P(B1 and B2)}$ of picking up a blue marble in the next draw is $\frac{3}{28}$.


Therefore, option (b.) is the correct answer.