Mathematics, 24.09.2019 01:00 Tweektweak

According to a survey, 60 % of the residents of a city oppose a downtown casino. of these 60 % about 8 out of 10 strongly oppose the casino. complete parts (a) through (c). (a) find the probability that a randomly selected resident opposes the casino and strongly opposes the casino. (b) find the probability that a randomly selected resident who opposes the casino does not strongly oppose the casino. (c) would it be unusual for a randomly selected resident to oppose the casino and strongly oppose the casino? explain.

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Answer from: RickyGotFanz4867

(a) 0.48

(b) 0.20

Step-by-step explanation:

(a) Find the probability that a randomly selected resident opposes the casino and strongly opposes the casino.

The probability that a radomly selected resident opposes the casino and strongly opposes the cassino is the product of the two probabilities, that a resident opposes the casino and that it strongly opposes the casino (once it is in the first group) as it is shown below.

Use this notation:

Probability that a radomly selected resident opposes the casino: P(A)Probability that a resident who opposes the casino strongly opposes it: P(B/A), because it is the probability of event B given the event A

i) Determine the probability that a radomly selected resident opposes the casino, P(A)

Probability = number of favorable outcomes / number of possible outcomes

P(A) is given as 60%, which in decimal form is 0.60

ii) Next, determine,the probability that a resident who opposes the casino strongly opposes it, P(B/A):

It is given as 8 out of 10 ⇒ P(B/A) = 8/10

iii) You want the probability of both events, which is the joint probability or intersection: P(A∩B).

So, you can use the definition of conditional probability:

P(B/A) = P(A∩B) / P(A)

iv) From which you can solve for P(A∩B)

P(A∩B) = P(B/A)×P(A) = (8/10)×(0.60) = 0.48

(b) Find the probability that a randomly selected resident who opposes the casino does not strongly oppose the casino.

In this case, you just want the complement of the probability that a radomly selected resident who opposes the casino does strongly oppose the casino, which is 1 - P(B/A) = 1 - 8/10 = 1 - 0.8 = 0.2.

You are being asked about the joint probability (PA∩B), which you found in the part (a) and it is 0.48.

That is almost 0.50 or half of the population, so you conclude it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the casino.

Answer from: Quest

is there any photos or whatever?

Answer from: Quest

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