a poll was conducted asking 560 if they used illegal drugs for recreation. the results were 270 answering yes, the rest answering no. if 10 people are selected at random, find the probability that 3 will have answered yes.
I haven't done this stuff in a while, forgot how to proceed, can someone give some guidelines and show how its done? I'm thinking I need to put 270 on one line, 290 no's on another, P(10...... yeah I'm lost.
I haven't done this stuff in a while, forgot how to proceed, can someone give some guidelines and show how its done? I'm thinking I need to put 270 on one line, 290 no's on another, P(10...... yeah I'm lost.
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P^x*q^n-x*C(nx) = prob of exactly x success = 0.134332
Because this is yes/no; pass/fail; success/failure type, Use Binomial Method:
where
n = the number of trials
x = probability of (0, or 1 or 2, up to n)
p = the probability of success in a single trial
q = the probability of failure in a single trial
(i.e. q = 1 − p)
Cnx is a combination
Here we have:
N= 10 P= 270/560 = 0.482143 Q=1- 0.482143 =0.51785714 ; C(nx) =120
x= 0.134332 or 13.43%
see source for a good study of this method
Because this is yes/no; pass/fail; success/failure type, Use Binomial Method:
where
n = the number of trials
x = probability of (0, or 1 or 2, up to n)
p = the probability of success in a single trial
q = the probability of failure in a single trial
(i.e. q = 1 − p)
Cnx is a combination
Here we have:
N= 10 P= 270/560 = 0.482143 Q=1- 0.482143 =0.51785714 ; C(nx) =120
x= 0.134332 or 13.43%
see source for a good study of this method
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Binomial Distribution with n = 560, p = 270/560, (1-p) = 290/560
Sample = 10
P(X = 3) = 10Choose3 * (270/560)^3 * (290/560)^7
10Choose3 = (10 * 9 * 8) / (3 * 2 * 1)
Sample = 10
P(X = 3) = 10Choose3 * (270/560)^3 * (290/560)^7
10Choose3 = (10 * 9 * 8) / (3 * 2 * 1)