A bag contains 6 blue balls and 3 green balls, and a box contains 4 blue balls and 5 green balls . Find the probability that a ball selected from either of them is a blue ball.
option
a) 1/3
b) 2/9
c) 5/9
d) 5/18

Either you can take the first bag with probability 1/2 and then pick a blue ball with probability 6/9 or you can take the second bag with probability 1/2 and pick a blue ball with probability 4/9. So the equation becomes
total probability= ((1/2)*(6/9)) + ((1/2)*(4/9)) = 5/9
10 years agoHelpfull: Yes(2) No(0)
for bag...
n(s)=9
n(e)=6
=>p1(e)=6/9
similerly for box,p2(e)=4/9
here ball is selected from either of one from two
so P(E)=(p1+p2)/2=5/18
10 years agoHelpfull: Yes(1) No(2)
we have 2 probability here
adding them we will get the ans

1st there are 6blue+3green =9 balls in bag
so probability getting blue ball here is 6c1/9c1

2nd there are 4blue+5green=9balls
so probability getting blue ball here is 4c1/9c1

adding them we will get 10/9

but we have to take 1 ball either one of them so divide the ans by 2
we will get the final ans as 5/9
10 years agoHelpfull: Yes(1) No(0)
Ans is 5/18
10 years agoHelpfull: Yes(0) No(1)

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