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In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random. What is the probability that at least one ball is of green colour ?
- A$\frac{4}{5}$
- B$\frac{3}{5}$
- C$\frac{1}{5}$
- D$\frac{2}{5}$
Answer not set
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0
CCommunity
Solution
There are 4 white, 4 red and 2 green balls and two balls are drawn at random.
Total possible outcomes = Selection of 2 balls out of 10 balls
= $C^{10}_2 = \frac{10 * 9}{1 * 2} = 45$
Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls
= $C^2_1 \times C^8_1 + C^2_2$
= 2*8 + 2 = 18
$\therefore$ Required probability = $\frac{18}{45} = \frac{2}{5}$
There are 4 white, 4 red and 2 green balls and two balls are drawn at random.
Total possible outcomes = Selection of 2 balls out of 10 balls
= $C^{10}_2 = \frac{10 * 9}{1 * 2} = 45$
Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls
= $C^2_1 \times C^8_1 + C^2_2$
= 2*8 + 2 = 18
$\therefore$ Required probability = $\frac{18}{45} = \frac{2}{5}$
0
CCommunity
Solution
There are 4 white, 4 red and 2 green balls and two balls are drawn at random.
Total possible outcomes = Selection of 2 balls out of 10 balls
= $C^{10}_2 = \frac{10 * 9}{1 * 2} = 45$
Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls
= $C^2_1 \times C^8_1 + C^2_2$
= 2*8 + 2 = 18
$\therefore$ Required probability = $\frac{18}{45} = \frac{2}{5}$
There are 4 white, 4 red and 2 green balls and two balls are drawn at random.
Total possible outcomes = Selection of 2 balls out of 10 balls
= $C^{10}_2 = \frac{10 * 9}{1 * 2} = 45$
Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls
= $C^2_1 \times C^8_1 + C^2_2$
= 2*8 + 2 = 18
$\therefore$ Required probability = $\frac{18}{45} = \frac{2}{5}$