Question 21656: I dont know where to begin with this, please help me!
After you complete this exam, you and two friends decide to go to McDonald’s to celebrate. Last month McDonald’s filled 90% of the orders accurately. What is the probability that:
a) All three orders will be filled accurately?
b) None of the three orders will be filled accurately?
c) At least 2 of the 3 orders will be filled accurately?
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! 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.
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