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Independent Events

Posted on 2016-07-21
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Can you please tell me if these calculations are correct?  Thanks.

A restaurant in a fast food franchise has determined that the chance a customer will order a soft drink is 0.95. The probability that a
customer will order a hamburger is 0.68. The probability that a customer will order french fries is 0.59.

a. If a customer places an​ order, what is the probability that the order will include a soft drink and no fries if these two events are​
independent? The probability is :  P(soft drink) = 0.95,         P(french fries) = 0.59, the complement of f(F) is 0.59

P(s intersection f), intersection of two independent events calls for multiplying each probability
                  P(s) * P(f) = 0.95 * 0.59 = 0.5605

b. The restaurant has also determined that if a customer orders a​ hamburger, the probability the customer will order fries is 0.72.
Determine the probability that the order will include a hamburger and fries.

The probability is  P(h intersection f) = 0.68 * 0.59 = 0.4012  The two events would not be independent.
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Question by:mustish1
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sdstuber earned 1000 total points
ID: 41723545
a - correct

b - incorrect

your multiplication is correct, but the chance of ordering fries goes up if the order also includes a hamburger.

So, the modified fries probability is still "independent" in that it is a distinct event even though that event is influenced by another event.
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Assisted Solution

by:Petr J
Petr J earned 1000 total points
ID: 41723569
a - IMHO it's incorrect. The calculation described above calculates probability of ordering soft drink and french fries, not soft drink without french fries.
Correct calculation:
P(s) * (1 - P(f)) = 0.95 * (1 - 0.59) = 0.3895

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Expert Comment

by:sdstuber
ID: 41723577
you are correct  the "A" answer IS wrong

I misread the question as "drink WITH fries"  not "drink WITHOUT fries"
I didn't see the "no"
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Author Closing Comment

by:mustish1
ID: 41723666
Thank You.
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