A box contains 6 red marbles numbered from 1 | vedclass

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(B) The total number of red marbles is $6$ (numbered $1, 2, 3, 4, 5, 6$).The total number of white marbles is $4$ (numbered $12, 13, 14, 15$).The tota

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$ \frac{1}{5} $

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(B) The total number of red marbles is $6$ (numbered $1, 2, 3, 4, 5, 6$).The total number of white marbles is $4$ (numbered $12, 13, 14, 15$).The total number of marbles in the box is $6 + 4 = 10$.We are looking for a marble that is both white and odd-numbered.The white marbles are ${12, 13, 14, 15}$.Among these,the odd-numbered marbles are $13$ and $15$.Thus,there are $2$ favorable outcomes.The probability $P$ is given by the ratio of favorable outcomes to the total number of outcomes:$P = \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{2}{10} = \frac{1}{5}$.

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

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