A lottery claims that 10% of tickets win a prize. How many tickets should you buy to be 50% sure of winning a prize?
This question is of binomial distribution/bernoulli trials. Can anyone help? I thought it might be:
solve for n in: nCr(n,r)*(0.1)^(r)*(.9)^(n-r) = .5
but I don't know what value to substitute for r.
Answer is 7.
Any help will be great. Thanks.
This question is of binomial distribution/bernoulli trials. Can anyone help? I thought it might be:
solve for n in: nCr(n,r)*(0.1)^(r)*(.9)^(n-r) = .5
but I don't know what value to substitute for r.
Answer is 7.
Any help will be great. Thanks.
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it is not binomial, it is hypergeometric, and the q is defective because it hasn't given how many lottery tickets there are.
however, by reverse calculations, i find that 100 tickets were sold to get an ans of 7
that, too can't be done analytically, but by trial & error
the way to go about it is to calculate P[NOT win a prize] ≤ 0.5
Go on multiplying 90/100 *89/99 *88/98 *.......... until you get a figure ≤ 0.5
we find that this happens for 90/100 *89/99 *88/98 *87/97 *86/96 *85/95 *84/94 = 0.47
so 7 is the ans
BAD q !
however, by reverse calculations, i find that 100 tickets were sold to get an ans of 7
that, too can't be done analytically, but by trial & error
the way to go about it is to calculate P[NOT win a prize] ≤ 0.5
Go on multiplying 90/100 *89/99 *88/98 *.......... until you get a figure ≤ 0.5
we find that this happens for 90/100 *89/99 *88/98 *87/97 *86/96 *85/95 *84/94 = 0.47
so 7 is the ans
BAD q !