A box contains $6$ red marbles numbers from $1$ through $6$ and $4$ white marbles $12$ 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 $\frac{1}{5}$
Number of white marbles are $4$ namely, $12, 13, 14, 15$
Now, the odd numberd marbles are $13, 15$ , two odd numbered white marbles
Total number of marbles are $6$ red marbles $+$ $4$ white marbles $= 10$ marbles
Hence, $P = \frac{N u m b e r o f w h i t e a n d o d d n u m b e r e d m a r b l e s}{T o t a l m a r b l e s} = \frac{2}{10} = \frac{1}{5}$

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

The correct option is B 15Number of white marbles are 4 namely, 12,13,14,15Now, the odd numberd marbles are 13,15, two odd numbered white marbles Total number of marbles are 6 red marbles + 4 white marbles =10 marblesHence, P=NumberofwhiteandoddnumberedmarblesTotalmarbles=210=15

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