Answer:
it has a frequency of 5
Step-by-step explanation:
Step-by-step explanation:
The data below is what was provided in the question and it is what I solved the question with
P(A1) = 0.23
P(A2) = 0.25
P(A3) = 0.29
P(A1 n A2 ) = 0.09
P(A1 n A3) = 0.11
P(A2 n A3) = 0.07
P(A1 n A2 n A3) = 0.02
a
P(A2|A1) = P(A1 n A2)/P(A1)
= 0.09/0.23
= 0.3913
We have 39.13% confidence that event A2 will occur given that event A1 already occured
b.)
P(A3 n A3|A1) = P(A2 n A3)n A1)/P(A1)
= 0.02/0.23
= 0.08695
We have about 8.7% chance of events A2 and A3 occuring given that A1 already occured.
C.
P(A2 u A3|A1)
= P(A1 n A2)u(A1 n A3)/P(A1)
= P( A1 n A2) + P(A1 n A3) - P(A1 n A2 n A3) / P(A1)
= (0.09+0.11-0.02)/0.23
= 0.18/0.23
= 0.7826
We have 78.26% chance of A2 or A3 happening given that A1 has already occured.
Step-by-step explanation;
ANSWER=C=-160/9f+5/9f
Answer:
Step-by-step explanation:
