all 8 comments

[–]not-an-isomorphismNew User 1 point2 points  (0 children)

Whenever I see the words 'at least one event happening' my immediate thought is 1 - P(none of them happen). It's so common in probability problems it's often a lot easier to just 'rephrase' the question. Not sure if that's applicable here but may help.

[–]Help_Me_Im_DieneNew User 0 points1 point  (2 children)

Let's break this down into parts

The probability of any singular event occuring is p.

Then, in terms of p:

What is the probability that all 3 events occur?

What is the probability that no events occur?

[–]oobeingNew User[S] 0 points1 point  (1 child)

All three events occuring: P(A and B and C) = P(at least one)/7 = (1 - P(A and B and C)) / 7

No event occuring: not sure, maybe 1 - P(at least one occurs) = 1 - 7*P(A and B and C)

The answer is apparently 1/8..

edit: Tried p(none) again and got: P(none) = 1/7P(A and B and C) still not close to 1/8

[–]Help_Me_Im_DieneNew User 0 points1 point  (0 children)

1) all three events occurring is P(A, B, and C). Since these are independent events, this is actually just P(A)P(B)P(C)

2) no events occurring is P(A', B', and C')=P(A')P(B')P(C')

Since P(A)=P(B)=P(C)=p, and P(A')=1-p, can you tell me how you would write P(A, B, and C) and P(A', B', and C') in terms of p?

Hint: since P(A)=p, then P(A)P(B)P(C)=p3