Aptitude - Probability - Discussion

@Nandini & @Monica.

In example clearly mentioned that the ball ins neither red nor green.

i.e blue.

We have 7 blue balls.

Therefore n(E) = 7.

Sample= 8+7+6=21.

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

Not selection of Red and Green. Simple subtracts 1 from 2/3.

1 - 2/3= 1/3 (Answer).

Please all students read question carefully what he ask in question. He said that 8 red, 6 green, 7 blue total=21 probability ?

He said neither from red and green so only one option is 7 blue.

7c1 which is n(a)=7 and n(S)=21c1=21

Formula is prob. p(A)=n(A)/n(S)

7/21 = 1/3.

Total number 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.
P(E) = 8/21.

What is the meaning that die is fair or not?

There is a die with 10 faces. It is not known that fair or not. 2 captains want to toss die for batting selection. What is the possible solution among the following?
a) If no. is odd it is head, if no. is even it is tail
b) If no. is odd it is tail, if no. is even it is head
c) Toss a die until all the 10 digits appear on top face. And if first no. in the sequence is odd then consider it as tail. If it is even consider it as head.
what is the exact solution

Hear there are 1 randomly selected means
green 8 blue 6
average of 2colors
so 8+6=14
14/2=7

Hi aruna I think you didn't read the question properly it is clearly mentioned that neither red nor green, so the remaining are blue. Hence we hve to consider blue balls. Hope you got it :).

Neither red nor blue.

Then why we are take 7 blue balls only.

Can't understand.

Please clear my doubt.

n(E) =7 because there are 7 blue balls.