Consider randomly selecting a student at a large university. let a be the event that the selected student has a visa card, let b be the analogous event for mastercard, and let c be the event that the selected student has an american express card. suppose that p(a) = 0.6, p(b) = 0.4, and p(a ∩ b) = 0.3, suppose that p(c) = 0.2, p(a ∩ c) = 0.12, p(b ∩ c) = 0.1, and p(a ∩ b ∩ c) = 0.07. (a) what is the probability that the selected student has at least one of the three types of cards?

0.75

Step-by-step explanation:

Given,

P(A) = 0.6, P(B) = 0.4, P(C) = 0.2,

P(A ∩ B) = 0.3, P(A ∩ C) = 0.12, P(B ∩ C) = 0.1 and P(A ∩ B ∩ C) = 0.07,

Where,

A = event that the selected student has a Visa card,  

B = event that the selected student has a MasterCard,  

C = event that the selected student has an American Express card,

We know that,

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

= 0.6 + 0.4 + 0.2 - 0.3 - 0.12 - 0.1 + 0.07

= 0.75

Hence, the probability that the selected student has at least one of the three types of cards is 0.75.

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

my answer is$ 12

step-by-step explanation:

because 15 -2=12