Could you clarify how to find Quantity A? It seems to me that the 0.06 you calculate is just the probability of A and B both happening, and doesn't take into account the fact that Quantity A is both OR either. Should quantity A be: P(a)*(1-P(b)) + P(b)*(1-P(a)) + P(a)*P(b) = 0.14+0.18+0.06=0.38 ?

I am quite confused like you.
According to my understanding P(either or both happening) = 1 - P (neither happening)
= 1 - (0.7* 0.8) = 1 -.56 = 0.44

Another way to figure this out is P(A)+ P(B) - P(A*B) = .2+.3-.06 = .44


Maybe I am getting the wording of the question wrong. I don't understand

I consent with you as this sounds a bit logical than explanation above

You guys are right!

The Quantity A is indeed P(a)*(1-P(b)) + P(b)*(1-P(a)) + P(a)*P(b) = 0.14+0.24+0.06=0.44.
Fixed it!

BTW conway1521 got the formula right bikachu got both the formula and the value right! Nice team work




The whole distribution is P(a)P(b) + P(a)(1-P(b))+P(b)(1-P(a)) +(1-P(a))(1-P(b)) =1. Both occuring, A not B, B not A, both not occuring respectively.

The question says that probability of either or both happening.

Either means P(A or B) and both means P(A and B)

Since P(A or B) = P(A) + P(B) - P(A and B)

there is also an or between these two (either or both)
which means that we add these
so..

P(A or B) + P(A and B) = P(A) + P(B) - P(A and B) + P(A and B)

which equals P(A) + P(B) = 0.5

B

Since A is the complement of B, all you need to know is whether P(B) > .5. P(B) = .56, so answer B.

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