A box contains $5$ red and $4$ white marbles. Two marbles are drawn successively from the box without replacement and the second drawn marble drawn is found to be white. Probability that the first marble is also while is?

\frac{3}{8}

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\frac{1}{2}

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\frac{1}{3}

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\frac{1}{4}

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Solution

The correct option is A $\frac{3}{8}$
Given:
Box has $5$ red + $4$ white $= 9$ balls
Two marbles are drawn without replacement.
To find:
The probability that the first marble is also white.
The second ball drawn is white hence white ball left $= 3$ . And total ball left $= 8$
Hence the probability that the first marble is also white is $\frac{3}{8}$

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A box contains 5 red and 4 white marbles. Two marbles are drawn successively from the box without replacement and the second drawn marble drawn is found to be white. Probability that the first marble is also while is?

A box contains $5$ red and $4$ white marbles. Two marbles are drawn successively from the box without replacement and the second drawn marble drawn is found to be white. Probability that the first marble is also while is?