Aptitude - Probability - Discussion

Total number of balls = 8 + 7 + 6,
=21.

Probability not red or green= number of events/sample space.
So, 7÷21 = 1/3.

Thanks all for explaining.

@All.

Here, You can solve it in two ways.

Method 1:
Total number of balls = 7+8+6= 21.
Neither red nor green, that means it is blue,
Number of blue = 7,
Probability of blue = number of blue balls÷ total number of balls.

Probability of blue balls= 7/ 21.

1/3.

Method 2:
Total number of balls = 21.
Number of Red = 8.
Prob of selecting red = 8/21 or.
Number of Green = 6.
Prob of selecting green = 6/21 or 2/7.
Neither Red nor Green = 1-(prob of red+ prob of green).
1 - (8/21 + 6/21).
1 - 14/21 = 7/21 or 1/3.

Neither red''️ nor green'' means (avoiding red and green) so we left out with blue means the one ball we have to pick up should be blue colour ball so.

P(E) = 7/21 = 1/3.

How n(E) =7?

How can p(e) =7?

Because neither red nor green clearly means its blue.

Total no.of balls =8+7+6=21,
Neither red nor green,
I.e.,we need to take 8,
Therefore p(e)=8/21.

Guys it may be a wrong answer.

The actual correct answer;

Total no.of balls=(8+7+6)=21.
let E = event that the ball drawn is neither red nor green.
=event that the ball drawn is red/

Therefore n(E) = 8 (given neither red nor green) make a note this point.
therefore p(E) = 8/21.

Probability of selection Red or Green= 8+6=14 / 21 = 2/3.

Then how 1/3?

How it is possible?