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Answer:
0.6,0.7,0.3 neither disjoint nor independent.
Step-by-step explanation:
Given that at a large university, 60% of the students have a Visa card and 40% of the students have a MasterCard.
A= visa card
B = Master card
P(A) = 0.60 and P(B) = 0.40
P(AUB)' = 0.30
i.e. P(AUB) = 0.70
Or P(A)+P(B)-P(AB) =0.70
P(AB)= 0.30
Randomly select a student from the university.
1) the probability that this student does not have a MasterCard.
2. the probability that this student has either a Visa card or a MasterCard.
=
3. Calculate the probability that this student has neither a Visa card nor a MasterCard.
=
4. Are the events A and B disjoint? Are the events A and B independent?
A and B have common prob 0.30 hence not disjoint.
P(AB) ≠P(A)P(B)
Hence not independent