In a large group of Americans in 2001-2002, the probability of visiting the emergency room once in the past 12 months was 0.14, and the probability of visiting twice was 0.04.

(a) What was the probability of visiting the emergency room one or two times?

(b) For two randomly chosen Americans in 2001-2002, what is the approximate probability that both visited the emergency room exactly once? (Round your answer to two decimal places.)

(c) Explain why we cannot solve for the probability that two people chosen from the same household both visited the emergency room exactly once.

Question

Lillian

Mathematics

14 April, 01:14

In a large group of Americans in 2001-2002, the probability of visiting the emergency room once in the past 12 months was 0.14, and the probability of visiting twice was 0.04.

(a) What was the probability of visiting the emergency room one or two times?

(b) For two randomly chosen Americans in 2001-2002, what is the approximate probability that both visited the emergency room exactly once? (Round your answer to two decimal places.)

(c) Explain why we cannot solve for the probability that two people chosen from the same household both visited the emergency room exactly once.

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Answers (1)

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Mckinley Ortega · Answer 1

A.) Let A be the event of the emergency room once and B be the event of visiting the emegency room twice, then

P (A or B) = P (A) + P (B) = 0.14 + 0.04 = 0.18

Therefore, the probability of visiting the emergency room one or two times is 0.18

b) P (A) and P (A) = 0.14 x 0.14 = 0.0196

Therefore, for two randomly chosen Americans in 2001-2002, the approximate probability that both visited the emergency room exactly once is 0.0196