Question

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11. A bag contains 3 red balls and 4 white balls. Two
balls are drawn one after the other without
replacement at random from of the bag. If the
second ball drawn in given to be white, what
is the probability that the st ball drawn is
also white?​

Answer 1

Answer:

the answer is

$\frac{5}{9}$

hope this helps you...

it is very easy hence I don't give you full explanation

Answer 2

Answer: 4/7

Step-by-step explanation:

There is some twist or trick involved in the question , which many will fail to notice and answer it incorrectly . Here's what it means :

Note that we only need the probability that the first ball is white . We also know that two balls are not drawn together , but are drawn one after the other . Keeping these things in mind we see that we actually do not need the information that the 2nd ball drawn is given to be white . Note that we are taking out the 1st ball first , then the 2nd ball . So when we are taking the 1st ball , the 2nd ball remains in the bag , so we have to include it in our probability as well (this is the only twist in the question) .

So the probability that the first ball is white = $\frac{(no. of possible white balls)}{total no. of balls}$

⇒ 4/7

I hope my answer helps you in clearing your doubts as well.