... each trial is p. Find the values of p if P(X=4)=0.12
Solutions and explanations would be very helpful! Thanks!
Solutions and explanations would be very helpful! Thanks!
-
Ok Merlin--- In this problem you're given everything except the probability of success, p
The formula for P(x) = nCx *p^x*(1-p) n-x.
n = 5, x = 4, and P(x) is 0.12. so 0.12 = 5!/(1!*4!)*p^4*(1-p). 0.12 = 5*p^4(1-p).
Now divide both sides by 5 and you have 0.024 = p^4 - p^5 and p must be in the interval (0,1).
You can solve this partially by graphical means and then by estimation.
There just happens to be two values of p in the interval; one at about 0.459 and the other at 0. 9732 approximately. I'm going to say the 0.459 is more logical.
The formula for P(x) = nCx *p^x*(1-p) n-x.
n = 5, x = 4, and P(x) is 0.12. so 0.12 = 5!/(1!*4!)*p^4*(1-p). 0.12 = 5*p^4(1-p).
Now divide both sides by 5 and you have 0.024 = p^4 - p^5 and p must be in the interval (0,1).
You can solve this partially by graphical means and then by estimation.
There just happens to be two values of p in the interval; one at about 0.459 and the other at 0. 9732 approximately. I'm going to say the 0.459 is more logical.
-
P(X= 4) = 5 p^4(1 - p) = 0.12
The values of p are the roots of this quintic equation.
The values of p are the roots of this quintic equation.