Aptitude - Probability - Discussion

1st ball = 5/7.
2nd ball = 4/6.
5/7 * 4/6 = 10/21.

P1st (not blue)= 1-P(blue) = 1-2/7 = 5/7.
P2nd(not blue)= 1-P(blue) = 1-2/6 = 4/6.
P1 and P2(not blue)=5/7 x 4/6 = 20/42 = 10/21.

The probability of the first ball not being blue is 5/7 and the probability of the second ball also not being blue is 4/6 and this is so because we have already drawn one ball and now the total number of balls got reduced to 6. Now where the 4 comes from? Its because we're assuming that the first ball we've drawn is a non-blue and therefore we have one less ball than 5 to draw from now.

I hope it was helpful.

A bag contains 2 red, 3 green and 2 blue balls, total = 7 balls.
(probability of choosing+probability of not choosing=1).

Let's consider the probability of choosing two balls is blue then,
probability of choosing P=(2/7)*(1/6) = 1/21.
so, the probability of not choosing (NP) = 1-P,
NP = 1-(1/21)
NP = 20/21,
I think 20/21 is the right answer.

Choose the first ball: There are 7 balls in total (2 red, 3 green, and 2 blue) , so you have 7 choices for the first ball.

Choose the second ball: After picking the first ball, there are 6 balls left for the second ball (since you're not concerned with color at this point).

Now, multiply the number of choices for the first and second balls together to get the total number of ways to draw any two balls: 7 choices × 6 choices = 42 ways.

Finally, calculate the probability that none of the balls drawn is blue by dividing the number of ways to draw without blue balls by the total number of ways to draw any two balls:

Probability = (Number of ways to draw without blue balls) / (Total number of ways to draw any two balls).
Probability = 20 / 42.

You can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:

Probability = (20 ÷ 2) / (42 ÷ 2).
Probability = 10 / 21.
So, the probability that none of the balls drawn is blue is 10/21.

Total Balls = 7.
Num of Selected Balls are 2.
So, the Total Outcomes = 7C2.
Num of Blue Balls are Selected = 5C2.
So, the Probability = 5C2/7C2 = 10/21.

@Saiteja

I also thought that 20/21 is correct, actually, you have only deducted the probability of both blue from 1, you have to also deduct at least one blue from 1 i.e. 10/21. So the final calculation will go 1 - 1/21 - 10/21 = 10/21. (Probability of not getting blue = 1 - the probability of both blue - probability of one blue)

Using conceptual method rather than going by formula gives a better feel for what is happening.

P1 = probability of 1st ball not blue = 5/7 as 5 are non-blue.
P2 = probability of 2nd ball not blue = 4/6 as 1 non-blue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.

5/7 is it right?

Anyone can explain it, please.

7 CHOOSE 2 can also be denoted as 7C2.
7 is the total number of distinct elements (n),
2 is the number of elements drawn or chosen at a time (k),
21 is the total number of possible combinations (C).
nCk = n! / (k! * (n - k)!)
7C2 =7! / (2! x 5!) = 21.