A box contains $6$ red marbles numbered from $1$ through $6$ and $4$ white marbles numbered from $12$ through $15.$ Find the probability that a marble drawn 'at random' is white and odd numbered.

5

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\frac{1}{5}

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6

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$\frac{1}{6}$

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Solution

The correct option is B $\frac{1}{5}$
Total number of marbles $= 6 + 4 = 10$
Marble which is white and odd numbered $= {13, 15}$
Required probability $= \frac{2}{10} = \frac{1}{5}$

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A box contains 6 red marbles numbered from 1 through 6 and 4 white marbles numbered from 12 through 15. Find the probability that a marble drawn 'at random' is white and odd numbered.

The correct option is B 15 Total number of marbles =6+4=10 Marble which is white and odd numbered ={13,15} Required probability =210=15

A box contains $6$ red marbles numbered from $1$ through $6$ and $4$ white marbles numbered from $12$ through $15.$ Find the probability that a marble drawn 'at random' is white and odd numbered.

The correct option is B $\frac{1}{5}$
Total number of marbles $= 6 + 4 = 10$
Marble which is white and odd numbered $= {13, 15}$
Required probability $= \frac{2}{10} = \frac{1}{5}$