Question

A box contains $6$ red and $4$ white marbles. A marble is drawn and replaced back in the box for three times. The probability that two white marbles are drawn is:

• A. $\frac{34}{125}$
• B. $\frac{36}{125}$
• C. $\frac{38}{125}$
• D. $\frac{32}{125}$

Answer 1

The correct option is B $\frac{36}{125}$

Total marbles $=6$ Red and $4$ white $=10$
Cases when two white marbles are drawn and $1$ Red marble is drawn, is:
$WWR,WRW,RWW$

Hence, probability that two white marbles are drawn
$=3\times \left(\frac{{}^4{\text{C}}_1}{{}^{10}{\text{C}}_1}\right)\times \left(\frac{{}^4{\text{C}}_1}{{}^{10}{\text{C}}_1}\right)\times \left(\frac{{}^6{\text{C}}_1}{{}^{10}{\text{C}}_1}\right)=\frac{36}{125}$