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# A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

**(i)** red?

**(ii)** white?

**(iii)** not green?

Given:

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random.

To do:

We have to find the probability that the marble taken out will be

(i) red

(ii) white

(iii) not green

Solution:

Number of red marbles $=5$

Number of white marbles $=8$

Number of green marbles $=4$

Total number of marbles $=5+8+4=17$

This implies,

The total number of possible outcomes $n=17$.

(i) Total number of favourable outcomes(taking out a red marble) $=5$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that the marble taken out will be red $=\frac{5}{17}$

The probability that the marble taken out will be red is $\frac{5}{17}$.

(ii) Total number of favourable outcomes(taking out a white marble) $=8$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that the marble taken out will be white $=\frac{8}{17}$

The probability that the marble taken out will be white is $\frac{8}{17}$.

(iii) The total number of possible outcomes $n=17$.

Here, not green implies red or white marbles.

Total number of favourable outcomes(taking out a red or white marble) $=5+8=13$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that the marble taken out will be not green $=\frac{13}{17}$

The probability that the marble taken out will be not green is $\frac{13}{17}$.