Question

4) As part of pharmaceutical testing for drowsiness as a side effect of a drug, 200 patients are randomly assigned to one of two groups of 100 each. One group is given the actual drug and the other a placebo. The number of people who felt drowsy in the next hour is recorded as Drowsy Not drowsy Drug 40 60 Placebo 20 80 a. What is the probability that a randomly picked patient in the study feels drowsy in the next hour? b. What is the probability that a randomly picked patient in the study takes the placebo or feels drowsy in the next hour? c. Given that the patient was given the drug, what is the probability that he or she feels drowsy in the next hour? d. Is whether a patient feels drowsy independent of taking the drug? Explain using probabilities

Answer 1

the necessary table:-

	drug	placebo	ROW TOTAL
drowsy	40	20	60
not drowsy	60	80	140
COLUMN TOTAL	100	100	200

a). the probability that a randomly selected patient feels drowsy in the next hour be:-

= total number of patients who feels drowsy / total number of patients.

$=\frac{60}{200}$

= 0.3

b). the probability that a randomly selected patient takes the placebo or feels drowsy in the next hour be:-

= total number of patients who takes placebo + total number of patients who feels drowsy - number of patients who takes placebo and feels drowsy / total number of patients.

$=\frac{100+60-20}{200}$

= 0.7

c). given that the patient was given the drug , the probability that he or she feels drowsy in the next hour be:-

= P( drowsy | drug )

= total number of patients who was given the drug and feels drowsy / total number of patients was given the drug.

$=\frac{40}{100}$

= 0.4

d). probability of taking drug = P(drug) = 100/200 = 0.5

probability that a patient feels drowsy = P(drowsy) = 60/200 = 0.3

the two events are said to be independent if,

$P(drug\cap drowsy)=P(drug)*P(drowsy)$

here,

$P(drug\cap drowsy)$

= number of patients who takes drug and feels drowsy / total number of patients

= 40 / 200

= 0.2

$P(drug)*P( drowsy)$

= 0.5 * 0.3

= 0.15

hence,

$P(drug\cap drowsy)\neq P(drug)*P(drowsy)$

so, we can say that whether a patient feels drowsy is not independent of taking the drug.

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