I have 5 indepentant situations. There probability of each situation occuring is as follows:
1 : 60%
2 : 70%
3 : 85%
4 : 90%
5 : 66%
With all that known what is the probability that exactly 0,1,2,3,4, or 5 occur. Also exaplain how you got to the answer, so I could repeat the process with different %'s.
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A1 = 0.6 B1 = 0.4
A2 = 0.7 B2 = 0.3
A3 = 0.85 B3 = 0.15
A4 = 0.9 B4 = 0.1
A5 = 0.66 B5 = 0.34</pre><hr />
A's are probability of it happening, B's are probability of it not happening.
A's are independent, so B's are independent. P(5) is product of the A's, and P(0) is product of the B's.
P(5) = A1*A2*A3*A4*A5 = 21.21%
P(0) = B1*B2*B3*B4*B5 = 0.06%
P(1) is sum of the probabilities of each event happening times the probabilities of each other event not happening.
P(1) =
A1*B2*B3*B4*B5 + A2*B1*B3*B4*B5 + A3*B1*B2*B4*B5 +
A4*B1*B2*B3*B5 + A5*B1*B2*B3*B4
= 1.25%
P(4) is sum of the probabilities of each event not happening times the probability of each other event happening. Just reverse A's and B's in P(1).
P(4) =
B1*A2*A3*A4*A5 + B2*A1*A3*A4*A5 + B3*A1*A2*A4*A5 +
B4*A1*A2*A3*A5 + B5*A1*A2*A3*A4
= 40.25%
P(2) = Sum of probabilities of each pair of events happening times the probabilities that other 3 events do not happen.
P(2) =
A1*A2*B3*B4*B5 + A1*A3*B2*B4*B5 + A1*A4*B2*B3*B5 + A1*A5*B2*B3*B4 + A2*A3*B1*B4*B5 + A2*A4*B1*B3*B5 + A2*A5*B1*B3*B4 + A3*A4*B1*B2*B5 + A3*A5*B1*B2*B4 + A4*A5*B1*B2*B3
= 8.97%
=8.97%
For P(3) you can reverse A's and B's in P(2), or just take
P(3) = 1 - [ P(1) + P(2) + P(3) + P(5) ] = 28.26%
I did this by reversing A's and B's as a check, and got the same thing.