Mathematics, 06.10.2019 09:01, sloane50

Suppose that in a senior college class of 500500 students, it is found that 215215 smoke, 252252 drink alcoholic beverages, 215215 eat between meals, 124124 smoke and drink alcoholic beverages, 7878 eat between meals and drink alcoholic beverages, 9191 smoke and eat between meals, and 4646 engage in all three of these bad health practices. if a member of this senior class is selected at random, find the probability that the student (a) smokes but does not drink alcoholic beverages; (b) eats between meals and drinks alcoholic beverages but does not smoke; (c) neither smokes nor eats between meals.

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answered: darrenturner

The probability that the student smokes but does not drink alcoholic beverages is 0.182.

The probability that the student eats between meals and drinks alcoholic beverages but does not smoke is 0.064

The probability that the student neither smokes nor eats between meals is 0.322

Step-by-step explanation:

Consider the provided information.

Suppose that in a senior college class of 500 students, it is found that 215smoke, 252 drink alcoholic beverages, 215 eat between meals, 124 smoke and drink alcoholic beverages, 78 eat between meals and drink alcoholic beverages, 91 smoke and eat between meals, and 46 engage in all three of these bad health practices.

Let A is the represents student smoke.

B is the event, which represents student drink alcoholic beverage.

Part (A) Smokes but does not drink alcoholic beverages.

From the above information.

P(A) = 215/500 and P(A ∩ B) = 124/500

Thus,

$P(A\cap B')=P(A)-P(A\cap B)$

$P(A\cap B')=\frac{215}{500}-\frac{124}{500}=0.43-0.248=0.182$

The probability that the student smokes but does not drink alcoholic beverages is 0.182.

Part (B) Eats between meals and drinks alcoholic beverages but does not smoke.

Let C is the event, which represents student eat between meals.

Eats between meals and drinks alcoholic beverages but does not smoke can be written as:

$P(C\cap B \cap A')=P(B\cap C)-P(A\cap B \cap C)$

From the given information.

P(B ∩ C) = 78/500 and P(A ∩ B ∩ C)= 46/500

Thus,

$P(C\cap B \cap A')=\frac{78}{500}-\frac{46}{500}=0.156-0.092=0.064$

The probability that the student eats between meals and drinks alcoholic beverages but does not smoke is 0.064

Part (C) Neither smokes nor eats between meals.

Neither smokes nor eats between meals can be written as:

$P((A\cup C)')=1-P(A\cup C)$

From the given information.

P(A) = 215/500 and P(C)= 215/500 and P(A∩C)=91/500

P(A∪C)=P(A)+P(C)-P(A∩C)

P(A∪C)= $\frac{215}{500}+\frac{215}{500}-\frac{91}{500}=\frac{339}{500}$

Substitute the respective values in the above formula.

$P((A\cup C)')=1-P(A\cup C)=1-\frac{339}{500} \\1-0.678=0.322$

The probability that the student neither smokes nor eats between meals is 0.322

answered: Guest

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

answered: Guest

a .4y4

hope this !