Question

Q8. An urn contains 9 blue, 7 white and 4 black balls. If 2 balls are drawn at random, then what is the probability that only one ball is white?

Answer 1

Step-by-step explanation:

In total, there are 20 balls.

Let the event of drawing a white ball be event A

Let the event of drawing any ball other than white ball be event B

Now, probability of an event is the number of favourable outcomes divided by the total number of outcomes.

Considere probability of event A happening first, we have,

No. Of favourable outcomes (In this case, the number of white balls) = 7

Total no. Of outcomes (total number of balls)= 20

Then, P(A) = 7/20

Now, consider the probability of event B happening,

No. Of favourable outcomes = 13

Total no. Of outcomes = 19 (because 1 white ball has been selected already)

P(B) = 13/19

For these events to occur simultaneously, we multiply their probabilities.

P(A).P(B) = (7/20)*(13/19)

P(A).P(B) = 91/380

Also, there are two ways of this happening,

Event A can occur first and event B second, or Event B can occur first and event B second.

So, we multiply by 2.

The answer is 91/190

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