difficult problem?

Lets say,

Milk = M
Chicken = C
Apple = A

MAC = intersection of Milk, chicken and apple

similarly,
MA = intersection of Milk and apple
MC = intersection of Milk, chicken
AC = intersection of chicken and apple

then

100 = M + A + C - (MA+MC+AC) + MAC
(MA+MC+AC) = 60+50+35-100+10 = 45

We need to find all cases where only 2 items are purchased = (MA+MC+AC) - 2 x MAC = 45 - 2 x 10 = 25%

netigen wrote:Lets say,
We need to find all cases where only 2 items are purchased = (MA+MC+AC) - 2 x MAC = 45 - 2 x 10 = 25%

3. No of persons in exactly two of the sets: P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

Should it be 15?

I stand corrected:

No of persons in exactly two of the sets: P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)
No of persons in two or more of the sets: P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC)

so answer should be 15%

60+50+35-100+10 = 55

Hence 55-3X10 = 25

Am I correct?

dzelkas wrote:I can't seem to figure out the below problem. Can someone please help
the correct answer should be 25%

Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase 2 of the above products?



a) 5%
b) 10%
c) 15%
d) 25%
e) 30%

Basically, if someone is in two groups they're counted twice, so we need to subtract them once; if someone is counted in three groups they're counted three times, so we need to subtract them twice.

So:

True # = total group a + total group b + total group c - (ab + ac + bc) - 2(abc)

100 = 60 + 50 + 35 - (doubles) - 2(triples)

100 = 145 - 2(10) - doubles
doubles = 145 - 20 - 100
doubles= 145 - 120 = 25

Note that if there were also some people in none of the 3 groups, the formula would have been:

True # = total group a + total group b + total group c - (ab + ac + bc) - 2(abc) + total in none of a/b/c

but in this question we know that every shopper buys at least one product.