Question

A bag contains $8$ blue tickets and $12$ red tickets. $3$ tickets are drawn at random.What is the probability of getting all three blue tickets (with replacement)?

• A. $\frac{4}{25}$
• B. $\frac{8}{25}$
• C. $\frac{8}{125}$
• D. $\frac{4}{125}$

Answer 1

The correct option is C $\frac{8}{125}$

Given:
Total number of tickets $=8+12=20$

Number of blue tickets $=8$

Number of red tickets $=12$

Probability of choosing first blue ticket is $\text{P(A)}=\frac{8}{20}=\frac{2}{5}$

Probability of choosing second blue ticket is $\text{P(B)}=\frac{8}{20}=\frac{2}{5}$

Probability of choosing third blue ticket is $\text{P(C)}=\frac{8}{20}=\frac{2}{5}$

The probability of getting all three blue tickets is $\text{P(A, B and C)}=\text{P(E)}=\text{P(A)}\times \text{P(B)}\times \text{P(C)}$
$\Rightarrow \text{P(E)}=\frac{2}{5}\times \frac{2}{5}\times \frac{2}{5}$
$\Rightarrow \text{P(E)}=\frac{8}{125}$

So, the probability of getting all three blue tickets is $\frac{8}{125}$.