Probability of happening 2 from 3 events

Question

This is my second lesson in my statistics class, and I'm trying to understand the following problem:

I have 3 events A,B,C $\ Pr(A∩B)=0.25\\ Pr(B∩C)=0.15\\ Pr(A∩C)=0.2\\ Pr(A∩B∩C)=0.08$

And I have to find probability of happening 2 events from A B C (3 events)

My solution: $\ Pr(A∩B)=Pr(A)Pr(B)\\ Pr(B∩C)=Pr(B)Pr(C)\\ Pr(A∩C)=Pr(A)Pr(C)\\ Pr(A∩B∩C)=Pr(A)Pr(B)Pr(C)\\ => \\ Pr(A)=Pr(A∩B∩C)/Pr(B)Pr(C)=0.08/0.15=0.53\\ Pr(B)=Pr(A∩B∩C)/Pr(A)Pr(C)=0.008/0.25=0.32\\ Pr(C)=Pr(A∩B∩C)/Pr(A)Pr(B)=0.08/0.2=0.40$

So from here I have Pr(A) Pr(B) and Pr(C), probability of happening two of three events is: $\Pr(D)=Pr(A)+Pr(B)+Pr(c)-Pr(A∩B∩C)$

the problem is at substitution $\Pr(D)=0.53+0.32+0.4-0.08 = 1.25 $

and is wrong from what I know because probability can be a value from [0,1]

What I'm doing wrong?

Answer 1

Hint: Draw a Venn diagram which will reveal to you that \begin{align} P(AB) &= P(ABC) + P(ABC^c)\\ P(BC) &= P(ABC) + P(A^cBC)\\ P(AC) &= P(ABC) + P(AB^cC)\\ \end{align} and since $P(\text{exactly}~2~\text{of}~A,B,C) = P(ABC^c)+P(A^cBC)+ P(AB^cC)$, we have that $$P(\text{exactly}~2~\text{of}~A,B,C) = P(AB)+P(BC)+P(AC) - \mathbf{3}(PABC).\tag{1}$$ No need to calculate $P(A)$, $P(B)$ etc. or to assume independence.

If you need $P(\text{at least}~2~\text{of}~A,B,C)$, just add $(PABC)$ to the right side of $(1)$ to get $$P(\text{at least}~2~\text{of}~A,B,C) = P(AB)+P(BC)+P(AC) - \mathbf{2}(PABC).\tag{2}$$

Answer 2

The probability of two of the three events happening would be

$\Pr(A^c∩B∩C) + Pr(A∩B^c∩C) + Pr(A∩B∩C^c)$,

since the three event are trivially disjoints.

I'm guessing the events are disjoints, making the calculations you made possible.

The result is indeed between 0 and 1, as expected.