According to a center for disease control, the probability that a randomly selected person has hearing problems is 0.1530.153. The probability that a randomly selected person has vision problems is 0.0830.083. Can we compute the probability of randomly selecting a person who has hearing problems or vision problems by adding these probabilities? Why or why not?

(A) No, because hearing problems and vision problems are events that are too similar to one another.

(B) No, because hearing and vision problems are not mutually exclusive. So, some people have both hearing and vision problems. These people would be included twice in the probability.

(C) Yes, because this is an application of the Addition Rule for Disjoint Events.

(D) Yes, because hearing and vision are two different senses, and therefore, they are two unique problems.

Question

Mathematics

Catherine Dunn

12 September, 08:52

According to a center for disease control, the probability that a randomly selected person has hearing problems is 0.1530.153. The probability that a randomly selected person has vision problems is 0.0830.083. Can we compute the probability of randomly selecting a person who has hearing problems or vision problems by adding these probabilities? Why or why not?

(A) No, because hearing problems and vision problems are events that are too similar to one another.

(B) No, because hearing and vision problems are not mutually exclusive. So, some people have both hearing and vision problems. These people would be included twice in the probability.

(C) Yes, because this is an application of the Addition Rule for Disjoint Events.

(D) Yes, because hearing and vision are two different senses, and therefore, they are two unique problems.

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Answers (1)

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Skinny Jeans · Answer 1

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Probability that a randomly selected person has hearing problems = 0.153

Probability that a randomly selected person has vision problems = 0.083

We need to check whether probability of randomly selecting a person who has hearing problems or vision problems by adding these probabilities.

Since they are disjoint events as the same person does not have both the problems at the same time.

So, we can calculate by adding both.

P (H∪V) = 0.153+0.083=0.236

Hence, Option 'c' is correct.