Aptitude - Probability - Discussion

Here.
No of out comes is 7c2 and no of events is 3c2+2c2+2c1. 3c1 then no of events possible is.
P (E) =n (E) /n (s) =11/21.
But we have to find here is none of the two ball is blue so we have to find.
P (~E) =1-P (E) =10/21.

Is there any other way to understand basic concept of probability?

Formula is probability is p(A) = n(A)/n(s), so we find n(s) = 7C2 = 21.

Next we find n(a) = 5C2 = 10 these amounts put in formula

p(A)=n(A)/n(S) Ans is 10/21

We have interested in 2 balls none in white so 2+5 not 2 white balls I think you understood

There is
2 Red
3 Green
2 Blue Balls
i.e. 5 Non blue balls and 2 Blue

1st ball can be picked from those 5 non blue balls so that none of them is blue.

Hence n(E)=5 & n(S)= 7
Hence probability is 5/7

The next ball will be picked from remaining 6 balls and 4 Non-Blue balls(Since one is already picked)

Hence Probability(2nd picked ball not blue) = 4/6

Hence
Total probability = (5/7) * (4/6) = 10/21.

Is there any way to determine that when we take combination to find probability and when we take direct favourable events/total no of events. Actually I am confused in determining where to take combination and where not.

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2 `
= (7 x 6) /(2 x 1)
= 21.

A: Event none of the balls draw is blue.

n(A): number of possible ways = 2C1*3C1+2C2*3C0+3C2*2C0
=2*3+1*1+3*1
=6+1+3
=10.
and

P(A)=n(A)/n(S)=10/21=1/2.

If you are not aware of combinations like 7C2 then give names to the balls like r1, r2, g1, g2, g3, b1 and b2.

So if you take two balls out of the bag then if we consider r1 comes then total combination would be r1r2, r1g1, r1g2.. and so on. That means, for r1 there are 6 combinations.

So we can conclude following table.

Ball combination
r1 6
r2 5
g1 4
g2 3
g3 2
b1 1
b2 0

So total number of combination = 6+5+4+3+2+1 = 21.
And combination using blue ball = 2+2+2+2+2+1 = 11.

So combination without using blue ball = 21-11 = 10.

So probability of the same = 10/21 :).

Is it with or without replacement?

In the same question what is the probability of taking 2 blue balls?

In the same question what is the probability of taking 2 blue balls?