Question

Find P(A or B or C) for the given probabilities. P(A=0.38, P(B)=0.28, P(C)=0.11 P(A and B)= 0.11, P(A and C)=0.02, P(B and C)=0.07 P(A and B and C)= 0.01

Answer 1

Attached is a ven diagram which may be helpful for this problem.

Each number is a probability. The sum of all the numbers is P(A or B or C).

To find these numbers, start in the middle and work your way out.

The middle is P(A and B and C), which is 1. Then simply subtract this from each of the larger intersections. 

Finally the parts not in an intersection. Only in A not B not C. You just subtract the numbers in A from P(A).  ---> 38 - 10 -1 - 1 = 26

Add them up and you get 58.

P(A or B or C) = 0.58

Answer 2

Answer:  P(A ∪ B ∪ C) = 0.58.

Step-by-step explanation:  For the three events A, B and C, the following probabilities are given :

$P(A)=0.38,~~P(B)=0.28,~~P(C)=0.11,~~P(A\cap B)=0.11,\\\\~~P(B\cap C)=0.07,~~P(A\cap C)=0.02,~~P(A\cap B\cap C)=0.01.$

We are to find the value of P(A ∪ B ∪ C).

From the laws of probability, we have

$P(A\cup B\cup C)\\\\=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\\\\=0.38+0.28+0.11-0.11-0.07-0.02+0.01\\\\=0.78-0.2\\\\=0.58.$

Thus, P(A ∪ B ∪ C) = 0.58.