# Probability Question. Events A and B....

#### pumbaa213

The question goes like this:

The events A and B are such that P(A)=0.6, P(B)=0.3 and P(A|B)=0.2. Find the probability that
(i) both A and B occur,
(ii) at least one of A and B occurs,
(iii) exactly one of A and B occurs.

I'm fine with part (i) but have no idea how to go about solving (ii) and (iii).
Could somebody help me with it? Please give some explanations! Thank you very much (Talking)

#### romsek

MHF Helper
The question goes like this:

The events A and B are such that P(A)=0.6, P(B)=0.3 and P(A|B)=0.2. Find the probability that
(i) both A and B occur,
(ii) at least one of A and B occurs,
(iii) exactly one of A and B occurs.

I'm fine with part (i) but have no idea how to go about solving (ii) and (iii).
Could somebody help me with it? Please give some explanations! Thank you very much (Talking)
ii) Pr[at least one of A and B occurs] = Pr[A or B] = P[A] + P - P[A and B]

iii) Pr[exactly one of A and B occurs] = Pr[(A and ~B) or (~A and B)]

see if you can finish it

1 person

#### HallsofIvy

MHF Helper
Another way to do this is to use a "Venn diagram". Since $$\displaystyle \frac{6}{10}= .6$$ and $$\displaystyle \frac{3}{10}= .3$$, imagine 100 objects. There will be 60 in A and 30 in B. We need to determine how many there are in the intersection of A and B. Saying P(A|B)= .2 means that, of all the objects in B, .2 of them are in A. That is, .2(60)= 12 of the 60 object in A and 12 of the 30 objects in B are in their intersection. That leaves 60- 12= 48 in A but not in B, and 30- 12= 18 in B but not A for a total of 48+ 18+ 12= 78 in either A or B or both (equivalently, 60+ 30- 12= 78). The "the probability that at least one of A or B" is 78/100= 0.78. There are 100- 78= 28 in neither A nor B so 28/100= .28.

1 person

#### pumbaa213

Thank you soo much!!!! I've got it!