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Mathematics 16 Online
OpenStudy (anonymous):

I give medalsss Colby is testing the effectiveness of a new allergy medication. There are 100 people with allergies in the study. Sixty-two patients received the allergy medication, and 38 other patients did not receive treatment. Fifty of the patients who received the medication reported no allergies at the end of the study. Twenty of the patients who did not receive medication reported no allergies at the end of the study. What is the probability that a patient chosen at random from this study took the medication, given that they reported no allergies? guys please helppp

OpenStudy (anonymous):

@kropot72 can you help me?

OpenStudy (kropot72):

|dw:1406919576328:dw| Do you understand the probability tree?

OpenStudy (anonymous):

no can you explain me

OpenStudy (anonymous):

you just divide the number of patients that were treated by the total right?

OpenStudy (anonymous):

@kropot72 you there?

OpenStudy (anonymous):

can you just tell me how to solve it? i appreciate what you are typing but please just give me a formula, i have to finish this in 10 minutes

OpenStudy (kropot72):

"Sixty-two patients received the allergy medication, and 38 other patients did not receive treatment." This is where the probabilities 0.62 and 0.38 come from. "Fifty of the patients who received the medication reported no allergies at the end of the study." 50/62 = 0.806. Also 1.000 - 0.806 = 0.194. "Twenty of the patients who did not receive medication reported no allergies at the end of the study." 20/38 = 0.526. Also 1.000 - 0.526 = 0.474. The probability that a patient was treated and had no allergies is given by: \[P(treated/no\ allergies)=0.62\times0.806=you\ can\ calculate\] The probability that a patient was untreated and had no allergies is given by: \[P(untreated/no\ allergies)=0.38\times0.526=you\ can\ calculate\] The required probability is given by: \[\large \frac{P(treated/no\ allergies)}{P(treated/no\ allergies)+P(untreated/no\ allergies)}\]

OpenStudy (anonymous):

so i plug in the information that you gave into this formula?

OpenStudy (kropot72):

Basically. However you need to calculate the two values of probability to plug in.

OpenStudy (anonymous):

okay let me see.

OpenStudy (anonymous):

can you explain me how to plug in the information into that formula?

OpenStudy (kropot72):

The probability that a patient chosen at random from this study took the medication, given that they reported no allergies is given by: \[\large \frac{(0.62\times0.806)}{(0.62\times0.806)+(0.38\times0.526)}=you\ can\ calculate\]

OpenStudy (anonymous):

.714???

OpenStudy (anonymous):

i think that's the answer

OpenStudy (kropot72):

You are correct.

OpenStudy (anonymous):

;) thank you,

OpenStudy (kropot72):

You're welcome :)

OpenStudy (anonymous):

you saved my live, I really appreciate it.

OpenStudy (anonymous):

Thankyou SO MUCH! It was correct :D

OpenStudy (kropot72):

You're welcome :)

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