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.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

the answer is (-4,-4)

hope this ! : d