# complement question..

• Nov 23rd 2009, 01:41 AM
transgalactic
complement question..
"the probability to buy a suite is 0.22
the probability to buy a shirt is 0.3
the probability to buy a tie is 0.28
the probability to buy a suite and shirt is 0.11
the probability to buy a suite and tie is 0.14
the probability to buy a tie and shirt is 0.28

what is the probability that the customer will not by anything?"

i was told that the probability that the customer will not by anything
equals the complement of the probability that he buys at least 1 item
why???

as i see it to solve it we need to do
(1-0.11)*(1-0.14)*(1-0.1)=0.68

why am i wrong?
• Nov 23rd 2009, 06:14 AM
statmajor

P(buying at least one thing) => $\displaystyle P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) - P(A \cap B \cap C)$

and P(ABC) = 0 (since it's not given, I'm assuming the customer can't buy all three for some odd reason).
• Nov 23rd 2009, 06:58 AM
transgalactic
why cant we do

those events are not connected we as i see it

why we cant do it this way?
• Nov 23rd 2009, 07:09 AM
statmajor

Sorry, misread your last post, the reason why you can't do that is because in this example $\displaystyle P(A \cap B \cap C) = P(A)P(B)P(C)$ is not true since the 3 events are not independent.
• Nov 23rd 2009, 07:09 AM
Plato
Quote:

Originally Posted by transgalactic
why cant we do
those events are not connected we as i see it
why we cant do it this way?

You can do that way as long as you do it correctly.
But the process is just as long. You still must use inclusion/exclusion.

You see $\displaystyle \mathcal{P}(A^cB^cC^c)\ne \mathcal{P}(A^c) \mathcal{P}(B^c) \mathcal{P}(C^c)$ unless the events are independent.
From the given you can see that the events are not independent.
• Nov 23rd 2009, 07:13 AM
transgalactic
why they are not independant?

buying a shirt and buying a tie are different things,they are not the same
you cant buy a tie and partialy buy a shirt with it
• Nov 23rd 2009, 07:15 AM
statmajor
P(buying a shirt and tie) = 0.28

If they were independent then P(buy a shirt)*P(buy a tie) = 0.28
• Nov 23rd 2009, 07:18 AM
transgalactic
AAAAAAAAAAAAA

i got it now thanks :)
• Nov 23rd 2009, 07:23 AM
transgalactic
how to calculate
$\displaystyle P(A \cap B \cap C)$
• Nov 23rd 2009, 07:27 AM
statmajor
I think it's 0 in this case since the probability of buying all three wasnt given (and the events are not independent). Otherwise, you can't really solve the problem.

But I could be wrong.
• Nov 23rd 2009, 07:29 AM
transgalactic
ohh i didnt look very good, it is given 0.06
thanks :)
• Nov 23rd 2009, 08:00 AM
transgalactic
in the next part of the question i am asked to find
what is the chance of buying exactly one item

i was told to calculate the chance of buying at least 2 items.

tie=t
shirt=s
suite=su
it is being translated as:
(t and s) or (t and su) or (s and su)