Suppose at a large university, 60% of the students have a visa card and 40% of the students have a mastercard. also, 30% of the students neither have 1 a visa card, nor have a mastercard. randomly select a student from the university. given that this student has a visa card, what is the probability that this student also has a mastercard1. calculate the probability that this student does not have a mastercard. 2. calculate 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?

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

sorry i need the points

step-by-step explanation:

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answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: