Sorry about the lateness of this question. I was a bit distracted this morning.
A doctor knows that the drug he is prescribing is about 30% effective. He currently has four patients on the drug coming in during the next week. He's hoping to hear positive results for three out of the four.
What are the chances that his hopes will be met?
BTW, Did you notice the SocialVibe widget on the sidebar? Please, if you have time, give it a whirl. And then give me some feedback on how it works, what you think of it, etc... Good or bad, I want to know
I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
Thursday, September 03, 2009
Hopes and Wishes
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The SocialVibe thing doesn't work for me. Clicked on it, a popup showed with a question which character are you, so I picked a random character. Then, what was supposed to be the next page suddenly turned out not to be, it hung.ReplyDelete
The answer? My guess is 3/10
My guess is a 2.7 % chance.ReplyDelete
This is based on the two possible outcomes that meet his criteria.
Probability of 3 patients recovering and one not recovering + probability of all 4 patients recovering.
p(3,1) + p(4,0)
[(30/100)(30/100)(30/100)(70/100)] + [(30/100)(30/100)(30/100)(30/100)] = 0.027
Since he want to hear positive results for 3 out of 4,ReplyDelete
I would say the probability will be 27/1000 i.e 2.7%
As Jolson and techie pointed out, the answer is very low. I won't re-explain it, but 2.7% is the answer.ReplyDelete
Yagami, Thanks for trying the widget. I tried it this morning and had the same problem. Hopefully they get it straightened out.ReplyDelete
Mike, I think the chances are higher than that. There are four ways for 3 out of 4 patients to have positive results (P=Positive, N=Negative: NPPP, PNPP, PPNP, PPPN) and one way for 4 patients to have positive results. Therefore the probability that at *least* 3 out of 4 patients have positive results is:ReplyDelete
(4)(3/10)(3/10)(3/10)(7/10) + (1)(3/10)^4
756/10000 + 81/10000 = 837/10000 = 8.37%
Oops... Last week definitely wasn't my week. You're right. The chances of 3 out of the 4 patients getting positive results is 8.37%. Still depressingly low, but a lot higher than my original answer.
The 16 possibilities are:
The four results where 3 out of 4 are positive are as you stated and the chances add up to 8.37%.
Thanks for the correction!