There are 3 yellow, 5 white and 7 green balls in a bag. A ball is drawn from the bag at random. The probability that the ball drawn is not white, is:
$\Big(\dfrac{1}{3}\Big)$
$\Big(\dfrac{2}{3}\Big)$
$\Big(\dfrac{3}{5}\Big)$
$\Big(\dfrac{1}{2}\Big)$

Question

There are 3 yellow, 5 white and 7 green balls in a bag. A ball is drawn from the bag at random. The probability that the ball drawn is not white, is: 1. 2. 3. 4.

KnowledgeBoat · Accepted Answer

Total number of outcomes = 3 yellow + 5 white + 7 green balls = 15

Let E be the event of not getting white balls.

The number of favorable outcomes to the event E = 3 yellow balls + 7 green balls = 10

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{10}{15} = \dfrac{2}{3}$

Hence, option 2 is the correct option.

Mathematics

There are 3 yellow, 5 white and 7 green balls in a bag. A ball is drawn from the bag at random. The probability that the ball drawn is not white, is:
$\Big(\dfrac{1}{3}\Big)$
$\Big(\dfrac{2}{3}\Big)$
$\Big(\dfrac{3}{5}\Big)$
$\Big(\dfrac{1}{2}\Big)$

Probability

Answer

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There are 3 yellow, 5 white and 7 green balls in a bag. A ball is drawn from the bag at random. The probability that the ball drawn is not white, is:
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$\Big(\dfrac{2}{3}\Big)$
$\Big(\dfrac{3}{5}\Big)$
$\Big(\dfrac{1}{2}\Big)$

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