Aptitude - Probability - Discussion

Shekar your explanation is good but you should have included the intersection part of it for easy understanding.

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.

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.

p(3) = {3,6,9,12,15,18}.
p(5) = {5,10,15,20}.
P(3u5) = p(3)+p(5).

I don't understand how it's come
= 6/20+4/20 ?

As per Addition Rule of Probability, we can have two set of events - Mutually Exclusive & Non-Mutually Exclusive..where Mutually Exclusive events don't occur simultaneously or together and reverse logic with Non-Mutually Exclusive events.

Having this theorem told in above para, now i come to the context:

We have 2 events,

Event A: Ticket picked is multiple of 3.
Event B: Ticket picked is multiple of 5.

Now P(A) = {3,6,9,12,15,18} = 6/20.
And P(B) = {5,10,15,20} = 4/20.

Since in P(A) & P(B) we could see 15 to be common, we could conclude that the events are Mutually Exclusive..so for such Mutually Exclusive events the Addition Rule of Probability goes like this:

P(AUB) = P(A) + P(B) - P(A n B)

Where, P(A n B) always looks for what is common in two events, which is 15 in this case, so P(A n B) = 1/20.

Therefore, P(ticket picked is multiple of 3 or 5) = 6/20 + 4/20 - 1/20 = 9/20 is the answer.

Please anyone help me for solving this problem:

Condition 1: Whenever a white ball + black ball is taken out a black ball is placed in.

Condition 2: Whenever 2 black balls are taken out one white ball is placed in.

Condition 3: Whenever 2 white balls are taken out one white ball is placed inside.

Suppose user gives this input:

Black balls: 5.

White balls: 4.

Output should be any of these:

White ball/black ball/undetermined.

ANS IS 9/20.

PROBABILITY = NUMBER OF FAVORABLE EVENTS/NUMBER OF TOTAL EVENTS.

Events of greeting multiple 3 of 5= 3, 6, 9, 12, 15, 18, 5, 10, 20.

P(E) = P(F)/P(E).

P(E) = 9/20.

Sample space p(s) = 20c1 = 20.

Events occurred p(e) = (3,6,9,12,15,18,5,10,15,20) = 10.

= (3,6,9,12,15,18,5,10,20)=9

Probability = p(e)/p(s).

= 9/20.

15 also should come in the list right?

Yes, 9/20 is the correct answer.