given p(a) = 0.072, p(b) = 0.180, and p(c) = 0.027, and that events a, b, and c are mutually exclusive, what is the p(a or b or c)?

Given P(A) = 0.072, P(B) = 0.180, and P(C) = 0.027, and that events A, B, and C are mutually exclusive,then P(A or B or C) is 0.279

Solution:

Given that probability of event A is P(A) = 0.072

Probability of event B is P(B) = 0.180

Probability of event C is P(C) = 0.027

Also event A, B and C are mutually exclusive.

Need to determine P( A or B or C )

For mutually exclusive events  

P( A or B or C ) = P(A) + P(B) + P(C)

= 0.072 + 0.180 + 0.027  = 0.279

Hence probability of occurrence of event A or event B or event C , where A , B and C are mutually exclusive events is 0.279.

4 adults and 12 babies

step-by-step explanation:

4 x 7 = 28

12 x 15 = 180

180 + 28 =208

4 x 3 = 12

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x