22% of the customers visiting the suit department of a certain store will purchase a suit, 30% will purchase a shirt and 28% will purchase a tie. 11% of customers purchase a shirt and a suit, 14% both a suit and a tie and 10% buy a shirt and a tie. Only 6% of customers buy all three items. Let A be the event that the customer purchases a suit, B the event that the customer purchases a shirt and C the event that the customer purchases a tie. If there are 1000 customers in the store one week, how many will purchase exactly one of these items

If there are 1000 customers in the store one week, how many will purchase exactly one of these items

1000 CUSTOMERS*28%=280

Step-by-step explanation:

A The event that a persons buys a suit  

B The event that a person buys a shirt  

C The event that a person buys a tie  

P(A)=  22%

P(B)=  30%

P(C)=  28%

P(AB)= 11%

P(AC)= 14%

P(BC)= 10%

P(ABC)= 6%

A u B u C Is the event that any item is bougth

AC u AC u BC Is the event that any two events occured

So the wanted probability is

P[(A u B u C )(AB u AC u BC)^c

P[(A u B u C )=P(AB)+ P(BC)+P(BC)

P[(A u B u C ) =0.22+0.30+0.28-0.11-0.14.-0.10+0.06

                      =0,51

0,51=+0,23+P[(A u B u C )(AB u AC u BC)^c

        =0,28

1000 CUSTOMERS*28%=280

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation: