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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 $\frac{13}{17}$.
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