A box contains $3$ red marbles and $2$ white marbles. A marble is drawn and replaced three times from the box. The probability that exactly one red marble is drawn is

3 / 5

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9 / 125

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36 / 125

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6 / 125

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Solution

The correct option is C $36 / 125$
Total balls are $=$ $3$ Red and $2$ White marbles $=$ $5$ 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 Red marble is drawn $=$ $3 \times (\frac{^{2} C_{1}}{^{5} C_{1}}) \times (\frac{^{2} C_{1}}{^{5} C_{1}}) \times (\frac{^{3} C_{1}}{^{5} C_{1}}) = \frac{36}{125}$

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A box contains 3 red marbles and 2 white marbles. A marble is drawn and replaced three times from the box. The probability that exactly one red marble is drawn is

The correct option is C 36/125Total balls are = 3 Red and2 White marbles = 5 ballsCase when two white marble is drawn = 2 White and 1 Red marbles WWR;WRW;RWWProbability that one Red marble is drawn = 3×(2C15C1)×(2C15C1)×(3C15C1)=36125

A box contains $3$ red marbles and $2$ white marbles. A marble is drawn and replaced three times from the box. The probability that exactly one red marble is drawn is