Question 1033400: Example: In a cafeteria, 80% of the customers order chips and 60% order buns. If 20% of those
ordering buns do not want chips, find the probability that two customers chosen at random,
(i) both order chips but not buns.
(ii) exactly one of them orders a bun only.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 80% order chips.
60% order buns.
20% of those ordering buns don't order chips.
that means that 80% of those ordering buns also order chips.
.8 * 60% = 48% order buns and chips.
you have:
80% order chips.
60% order buns.
48% order chips and buns.
if you subtract the 48% that order chips and buns from the ones that order chips and the ones that order buns, you are left with:
32% order chips only.
12% order buns only
48% order chips and buns.
since the total needs to be 100% of the customer, that leaves 8% that don't order either buns or chips as far as i can see.
if i'm correct, then the percentages break down as follows:
32% order chips only.
12% order buns only
48% order chips and buns.
8% don't order chips or buns.
the probability that 2 customers, chose at random order chips but no buns would therefore be .32 * .32 = .1024 = 10.24%.
the probability that exactly one of them orders a bun only would be .12 * .88 * 2 = .2112.
the multiplication by 2 is because the first person chosen could be the one that orders the bun only, or the second person chosen could be the one that orders the bun only. the probability is doubled because it could be either way.
this is the best i can do with this problem.
it at least gives you food for thought about how a problem like this might be analyzed.
if the correct answer turns out not to be as shown above, please let me know what the correct answer is so i can expand my knowledge as well.
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