In a box containing 18 balls of 3 different colours red, green and blue, if 10 are red and 4 are green, then what is the probability of getting a blue ball in the first attempt?

$\frac{1}{2}$

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$\frac{2}{9}$

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$\frac{1}{3}$

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$\frac{1}{4}$

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Solution

The correct option is B
$\frac{2}{9}$

Given,

Total number of balls = 18

The number of red balls = 10

The number of green balls = 4

Hence, the number of blue balls = 18 - (10+4) = 18 - 14 = 4

P(getting a blue ball) = $\frac{n u m b e r o f b l u e b a l l s}{t o t a l n u m b e r o f b a l l s}$ = $\frac{4}{18}$ = $\frac{2}{9}$

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In a box containing 18 balls of 3 different colours red, green and blue, if 10 are red and 4 are green, then what is the probability of getting a blue ball in the first attempt?

The correct option is B 29 Given, Total number of balls = 18 The number of red balls = 10 The number of green balls = 4 Hence, the number of blue balls = 18 - (10+4) = 18 - 14 = 4 P(getting a blue ball) = number of blue ballstotal number of balls = 418 = 29