60 people work on 3 projects: ss, pi or pd. 23 are on ss, 28 are on pi, & 26 are on pd. 7 for ss & pd, 12 for pi & pd, & 6 for all 3 proj. How many are only on ss & pi (i.e., not all 3 projects)?

Question

possible answers are 4, 5 or 6... I have no clue

Answer 1

we can solve this by applying the formula,

p(aubuc)=p(a)+p(b)+p(c)- p(a∩b)-p(a∩c)- p(b∩c)+p(a∩b∩c)

Given,

given

p(ss)=23,

p(pi)=28,

p(pd)=26,

p(ss∩pd)=7,

p(pd∩pib)=12

,p(ss∩pi∩pd)=6

and p(ss U pi U pd)=60

we need to find p(ss∩pi)

by applying these values in the formula

p(ss u pi u pb)=p(ss)+p(pi)+p(pd)- p(ss∩pi)-p(pi∩pd)- p(ss∩pd)+p(ss∩pi∩pd)

we get , 60=23+28+26+p(ss∩pi)-7-12- +6

60=64-p(ss∩pi)

p(ss∩pi)=4

Therefore there are 4 people who are working on SS and Pi

Answer 2

Doug, I’m with you: I get -2, which is unreasonable. I wonder whether there’s a typographical error.

Let’s label each combination of projects:

S: Ss alone

I: Pi alone

D: Pd alone

Si: (ss,pi) alone

Sd: (Ss,pd) alone

Id: (Pi,pd) alone

A: All three

I assume each number represents all combinations including the named combination.

for example, the first sentence says 23 are on Ss. To me that means all the different combinations that include Ss: Ss + Si + Sd + A

S + Si + Sd + A = 23

I + Si + Id + A = 28

D + Sd + Id + A = 26

similarly, the fourth sentence says 7 = Sd but it must also include all three (A), so I read that as 7 = Sd + A

Sd + A = 7

Id + A = 12

A= 6

Finally, we know there are 60 working on one or more projects

S + Si + Sd + I + Id + D + A = 60

It is easy to discover

Id = 6, Sd = 1 and D = 13

But after that, I’m forced to discover Si = -2.

Sorry Doug, I can’t help any further. If you have any questions, please ask.