SOLUTION: In a cafeteria students may order any combination of Chips, Sandwiches and Cold drink. The probability that a student chooses a cold drink is 0.45, sandwiches an

Let capital C be the set of students who chose chips.
Let capital S be the set of students who chose sandwiches.
Let capital D be the set of students who chose cold drinks.

Let small letters a through h represent the probabilities that a randomly
selected student is a member of the region that letter is in.



In a cafeteria students may order any combination of Chips, Sandwiches and
Cold drink. The probability that a student chooses: 

a cold drink is 0.45,

d+e+f+g = 0.45

sandwiches and chips 0.19, 

b+e = 0.19

cold drink and sandwiches 0.15, 

e+f = 0.15

cold drink and chips 0.25, 

e+d = 0.25

cold drink or sandwiches 0.6, 

b+c+d+e+f+g = 0.6

cold drink or chips 0.84, 

a+b+d+e+f+g = 0.84

cold drink or chips or sandwiches 0.9. 

a+b+c+d+e+f+g = 0.9

Find the probability that a student chooses all three.

e = ?


(1).         d+e+f+g   = 0.45
(2).     b    +e       = 0.19
(3).           e+f     = 0.15
(4).         d+e       = 0.25
(5).     b+c+d+e+f+g   = 0.6
(6).   a+b  +d+e+f+g   = 0.84
(7).   a+b+c+d+e+f+g   = 0.9
(8).   a+b+c+d+e+f+g+h = 1.0

Find h by subtracting (7) from (8). 
Find c by subtracting (6) from (7), etc. 
It's a matter of substituting and subtracting equations.



Answer:  the probability that a student chooses all three = e = 0.1

Edwin

In my solution, I will use the letters D, S and C  to coding drinks, sandwiches and chips, respectively.


So, we have an Universal set  U  of all possible combinations, and P(U) = 1.


We also have  given

    P(D) = 0.45
     
    P(S ∩ C) = 0.19

    P(D ∩ S) = 0.15

    P(D ∩ C) = 0.25

    P(D U S) = 0.6

    P(D U C) = 0.84

    P(D U C U S) = 0.9.


There is a REMARKABLE formula in elementary probability theory

    P(D U C U S) = P(D) + P(C) + P(S) - P(D ∩ S) - P(D ∩ C) - P(C ∩ S) + P(D ∩ C ∩ S)    (1)

which is valid for any subsets (sets of events) of the Universal set.


In this formula, many terms are given, but not all. The terms P(S) and P(D) are not given.

Had I knew these two terms, I would be in position to calculate the value of P(D ∩ C ∩ S) from the formula (1) MOMENTARILY.



So, my goal now is to find these terms   P(S) and P(D).



To find P(S),  I will use another basic formula of the elementary probability theory  P(D U S) = P(D) + P(S) - P(D ∩ S).

By substituting all given values, I get  0.6 = 0.45 + P(S) - 0.15;  it implies  P(S) = 0.6 - 0.45 + 0.15 = 0.3.



To find P(C),  I will use similar basic formula                                       P(D U C) = P(D) + P(C) - P(D ∩ C).

By substituting all given values, I get  0.84 = 0.45 + P(C) - 0.25;  it implies  P(C) = 0.84 - 0.45 + 0.25 = 0.64.



Now I substitute all the given and found values into the formula (1)

    0.9 = 0.45 + 0.3 + 0.64 - 0.19 - 0.15 - 0.2 + P(D ∩ C ∩ S),


which gives me the ANSWER

    P(D ∩ C ∩ S) = 0.9 - 0.45 - 0.3 - 0.64 + 0.19 + 0.15 + 0.2 = 0.1.

    My credo, as always, is this : the solution should be as simple as possible,

    or, if you want, as adequately simple as a problem is.