Can someone help solve for Y
1 answer:
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Answer:
a)0.08 , b)0.4 , C) i)0.84 , ii)0.56
Step-by-step explanation:
Given data
P(A) = professor arrives on time
P(A) = 0.8
P(B) = Student aarive on time
P(B) = 0.6
According to the question A & B are Independent
P(A∩B) = P(A) . P(B)
Therefore
&
is also independent
= 1-0.8 = 0.2
= 1-0.6 = 0.4
part a)
Probability of both student and the professor are late
P(A'∩B') = P(A') . P(B') (only for independent cases)
= 0.2 x 0.4
= 0.08
Part b)
The probability that the student is late given that the professor is on time
=
=
= 0.4
Part c)
Assume the events are not independent
Given Data
P = 0.4
= = 0.4
= 0.4 x P
= 0.4 x 0.4 = 0.16
= 0.16
i)
The probability that at least one of them is on time
= 1-
= 1 - 0.16 = 0.84
ii)The probability that they are both on time
P = 1 -
= 1 -
= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56