Aptitude - Probability - Discussion

I can't understand.

P (e) ={3, 6, 9, 12, 15, 18, 5, 10, 20}.

We multiplying p (e) with s then we don't want to multiply s={7, 8 to 18}.

Hi murugesh what you explained is clear. Thank you.

15 creates a big confusion.

Because my answer is 1/2...:(
p(3)={3,6,9,12,15,18}
p(5)={5,10,15,20}
P(3u5)=p(3)+p(5)
=6/20+4/20
=6+4/20
=10/20
=1/2

Thank you murugesh for clarifying the doubt.

Please give me logical solution of the following problem.

A lady has fine gloves and hats in her closet- 18 blue, 32 red, and 25 yellow. The lights are out and it is totally dark. In spite of the darkness, she can make out the difference between a hat and a glove. She takes out an item out of the closet only if she is sure that if it is a glove. How many gloves must she take out to make sure she has a pair of each color?

a) 50 b) 8 c) 60 d) 42

Thanks.

Hi
The answer is 9/20
A={3,6,9,12,15,18}
B={5,10,15,20}
P(A)=6/20
P(B)=4/20
A intersection B = {15}
P(A intersection B) = 1/20
P(A U B)= P(A)+P(B)-P(A intersection B)
= 6/20 + 4/20 - 1/20
= 9/20

A(n)={3,6,9,12,15,18}
B(n)={5,10,15,20}
P(AuB)=P(A)+P(B)-P(A intersection B)
P(AuB)=6/20+4/20-1/20
P(AuB)=9/20.

A(n)={3,6,9,12,15,18}
B(n)={5,10,15,20}
P(AuB)=P(A)+P(B)-P(A intersection B)
P(AuB)=6/20+4/20-1/20
P(AuB)=9/20.

Please how do you come about the 1/2 (ie A intersection B) ? I am a bit confused.
Thanks.

@Naana.

I will solve your problem.

->In set A there are 6 sample points and in set of B there are 4 sample points
->Now we want the intersection of two sets
->So we select the common sample points in both sets i.e 15
->Now in the set (A intersection B) we have only one sample point
->For finding probability of A intersection B
i.e p(A intersection B)

* So according to the definition of probability.

P(A intersection B)== n(A intersection B)/no of events;

So the result is P(A intersection B)=1/20.