Aptitude - Probability - Discussion

See there are two ways to solve this question:

METHOD 1:

Probability of getting no blue balls in 2 draws = probability of getting 2 other balls in 2 draws out of any other 5 balls.

i.e., select two balls from the remaining 5 balls in 5C2 ways = n(E),

Now the sample space is n(S) = selection of 2 balls out 7b balls = 7C2,

So, required probability= n(E)/n(S),

=> 5C2/7C2 = [5!/(2!*3!)]/[7!/(2!*5!)]

=> 10/21 is the answer.

METHOD 2:

Probability of not getting any blue balls in two draws = 1 - (probability of getting 1 blue ball and 1 other ball + probability of getting 2 blue balls)

So,

Selection of 1 blue ball and one other ball = 2C1 * 5C1 = 10 ways,

Selection of 2 blue balls = 2C2 = 1,

So, probability = 1 - [{10/7C2} + {1/7C2}]

=> 10/21 answer.

Could anybody tell me what is that c means in 7c2?

Why not the answer is 5/7?

What if we use the tree diagrams because the way you answer is kindly complicated.

Please clear me on how 6 come in?

Hey it should be 5/7.

Because if we subtract no of blue from total.

Then 7-(2+3) log%5 = 7*5*3*5*6 = log 7+5+3+6+1 = log 35 + log 15 + log 30.

So if we take log as common the it will be easy i.e = log (35+15+30).

= log*80 = 80 log the 80-75 = 5.

There fore the possibility is 5/7.

Why is E used?

How we can know that we need to use ''C'' formula in sum ?

Cannot we have a different method.

It will be like 1- putting out blue balls which is 2/7.

To probability will be 1-2/7 = 5/7?

There are two bags A and B. A contains n white and 2 black balls and B contains 2 white and n black balls. One of the two bags is selected at random and two balls are drawn from it without replacement.

If both the balls drawn are white and the probability that the bag A was used to draw the balls is 6/7, find the value of n.

Please answer with solution.