Complement question (statistics)

  • Thread starter Thread starter lom
  • Start date Start date
  • Tags Tags
    Statistics
lom
Messages
29
Reaction score
0
"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
not buy a shirt AND not buy a tie AND not buy suite=
(1-0.11)*(1-0.14)*(1-0.1)=0.68

why am i wrong?
 
Physics news on Phys.org
First, iN your approach you are assuming the choices are independent - they may not be.

Second, you are trying to find thechance JUST that group of 3 things; you haven't accounted for the purchase of a subset of them.

Think this way: either some item ispurchased or nothing is -those are complements.
 
of course the choises are independent
a shirt differs from a tie etc..

"you are trying to find thechance JUST that group of 3 things; you haven't accounted for the purchase of a subset of them. "
and i don't want to encounter their purchase
thats why i am doing complement

i still can't see my mistake in this way.

i tried to think your way
so i need the complement of
"either some item is purchased"
so we have 7 combinations
to buy some thing
_ _ _

for example :
to buy a tie and a suite but not to buy a shirt

etc..

why not
not buy a shirt AND not buy a tie AND not buy suite=
(1-0.11)*(1-0.14)*(1-0.1)=0.68thats alot
is there easier way?
 
lom said:
of course the choises are independent
a shirt differs from a tie etc..

"you are trying to find thechance JUST that group of 3 things; you haven't accounted for the purchase of a subset of them. "
and i don't want to encounter their purchase
thats why i am doing complement

i still can't see my mistake in this way.

i tried to think your way
so i need the complement of
"either some item is purchased"
so we have 7 combinations
to buy some thing
_ _ _

for example :
to buy a tie and a suite but not to buy a shirt

etc..

why not
not buy a shirt AND not buy a tie AND not buy suite=
(1-0.11)*(1-0.14)*(1-0.1)=0.68


thats alot
is there easier way?

IF the only two possibilities were to purchase all three items or to purchase nothing, your idea (not the calculation, as you don't know whether the choices to purchase the items are independent) would be correct.


However, there are more than two choices:

* purchase nothing
* purchase exactly one of the three items
* purchase exactly two of the three items
* purchase all of the items

Now do you see why, and where, your approach goes awry?
 
why you think that those events are linked??

buying a tie is not buying a shirt?
 
lom said:
why you think that those events are linked??

buying a tie is not buying a shirt?

You are confusing disjoint (events having nothing in common) with independent (a probability based idea).

The events "buy only a tie" and "buy only a shirt" are disjoint - there are no outcomes in common. THEY ARE NOT independent: if you tell me the person purchased a tie (and only a tie), then I know the probability they purchased a shirt is zero.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top