Question

A jar has $25$ yellow jelly beans and $15$ red jelly beans. If Samuel picks two jelly beans one by one without replacement at random, what is the probability that they are both red?

• A. $\frac{3}{52}$
  
• B. $\frac{7}{52}$
  
• C. $\frac{3}{8}$
  
• D. $\frac{14}{39}$
  

Answer 1

The correct option is B $\frac{7}{52}$


There are two events involved. These are dependent events as the picking up of one jelly bean affects the probability of other jelly bean being picked.

Event 1:
Number of red jelly beans $=15$
Total number of jelly beans $=40$

The probability $\text{P(R1)}$ that the first jelly bean picked up is red:

$\text{P(R1)}=\frac{\text{Number of red jelly beans}}{\text{Total number of jelly beans}}$

$\text{P(R1)}=\frac{15}{40}$

$\text{P(R1)}=\frac{3}{8}$



Event 2:
Now,
Number of red jelly beans left $=14$
Total number of jelly beans $=39$

The probability $\text{P(R2)}$ that the first jelly bean picked up is red:

$\text{P(R2)}=\frac{\text{Number of red jelly beans left}}{\text{Total number of jelly beans}}$

$\text{P(R2)}=\frac{14}{39}$



The probability $\text{P(Both R)}$ that the first jelly bean and the second jelly bean picked up both are red is obtained by multiplying $\text{P(R1)}$ and $\text{P(R2)}$

$\text{P(Both R)}=\text{P(R1)}\times \text{P(R2)}$

$\text{P(Both R)}=\frac{3}{8}\times \frac{14}{39}$

$\text{P(Both R)}=\frac{7}{52}$


The probability $\text{P(Both R)}$ that the first jelly bean and the second jelly bean picked up both are red is $\frac{7}{52}$.

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