Question 21656
P(fulfilling an order successfully) = 0.90 and
P(not...) = 0.10


So,
a) P(all 3 successful) = P(first OK) and P(second OK) and P(third OK)
--> 0.90 * 0.90 * 0.90
--> 0.729


b) P(none successful) = P(first not) and P(second not) and P(third not)
--> 0.10 * 0.10 * 0.10
--> 0.001


c) P(at least 2 of the 3) means a few things: It means...

1. P(first OK) and P(second OK) and P(third OK)
OR
2. P(first OK) and P(second OK) and P(third not)
OR
3. P(first OK) and P(second not) and P(third OK)
OR
4. P(first not) and P(second OK) and P(third OK)


1. is answer to a).
Next three versions (2, 3, 4) are each 0.90*0.90*0.10 --> 0.081.


As there are 3 different routes to get to it... 2, 3 or 4, so the probability is 3*0.081 --> 0.243.


So total probability of getting 2 or more orders correct is 0.729 + 0.243
--> 0.972


jon.