Question

A box contains 6 red and 4 white marbles. A marble is drawn and replaced three times from the box. The probability that two white marbles are drawn

• A. $34/125$
• B. $36/125$
• C. $38/125$
• D. $32/125$

Answer 1

The correct option is B $36/125$

Total balls are $=$ 6 Red and 4 White marbles $=$ 10 balls
Case when two white marble is drawn $=$ 2 White and 1 Red marbles
W W R; W R W; R W W
Probability that one white marble is drawn $=$ $3\times \left(\frac{{}^4{C}_1}{{}^{10}{C}_1}\right)\times \left(\frac{{}^4{C}_1}{{}^{10}{C}_1}\right)\times \left(\frac{{}^6{C}_1}{{}^{10}{C}_1}\right)=\frac{36}{125}$