If 10% of the people who take a certain medicine get a headache, find the probability that if 5 people take the medicine, one will get a headache.
2007-05-02 20:38:40 · 4 answers · asked by volleyballbabe814 1 in Science & Mathematics ➔ Mathematics
This is a standard binomial distribution with 5 independent trials and probability of success in each trial 0.1. The probability that exactly n people will get a headache is therefore C(5, n) . (0.1)^n . (0.9)^(5-n), where C(5, n) represents 5 choose n = 5! / n!(5-n)!.
P(exactly one gets a headache) = C(5, 1) . (0.1)^1 . (0.9)^4
= 5 (0.1) (0.6561) = 0.328 to 3 d.p.
P(at least one gets a headache) = 1 - P(none get one) = 1 - (0.9)^5 = 0.410 to 3 d.p.
2007-05-02 20:43:44 · answer #1 · answered by Scarlet Manuka 7 · 4⤊ 0⤋
If 10% of people will get a headache, then the probability is 1 in 10 or 0.1
so if 5 people take the medicine, and we know that 10% will get a headache, then the probability if 5 people take the medicine is 0.5 since 10% of 5 is 0.5
2007-05-02 20:50:47 · answer #2 · answered by Ratbag 2 · 1⤊ 3⤋
Is the question "at least 1" ? In that case it is easiest to calculate probability that nobody gets a headache and subtract from 1:
1 - 0.9^5 = about 0.41
If question is "exactly 1" then:
5*0.1*0.9^4 = 0.32805
2007-05-02 20:48:06 · answer #3 · answered by blighmaster 3 · 2⤊ 0⤋
No one... since 5 * .10 is .5..... And that's below a single person?
2007-05-02 20:45:08 · answer #4 · answered by Chris Lagmay 3 · 0⤊ 6⤋