Question 239414
<font face="Garamond" size="+2">


The probability of exactly *[tex \Large k] successes out of *[tex \Large n] trials is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k)\ =\ \left(n\cr k\right)\,p^k\,q^{n-k}]


Where *[tex \Large p] is the probability of success on a given trial, *[tex \Large q] is the probability of failure on a given trial, and *[tex \Large \left(n\cr k\right)] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time, or *[tex \Large \frac{n!}{k!(n-k)!}]


1.  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(4)\ =\ \left(4\cr 4\right)\,(0.9)^4\,(0.1)^0]


2.  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(0)\ =\ \left(4\cr 0\right)\,(0.9)^0\,(0.1)^4]


You get to do your own arithmetic.  Hints:  *[tex \Large \left(n\cr n\right)\ =\ \left(n\cr 0\right)\ =\ 1] and *[tex \Large a^0\ = 1]


3.  At least one does NOT arrive within 15 minutes.  That is the probability that none, 1, 2, or 3 do not arrive within 15 minutes -- which is *[tex \Large P_4(0)\ +\ P_4(1)\ +\ P_4(2)\ +\ P_4(3)].  That looks like a lot of calculation, but not really.  Since we know that either none or at least one must be late, then we can be certain that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(0)\ +\ P_4(1)\ +\ P_4(2)\ +\ P_4(3)\ +\ P_4(4)\ =\ 1]


But *[tex \Large P_4(4)] is the probability that NONE are late, so



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ P_4(4)]


is the probability that at least 1 is late.  This is a simple subtraction since you already have the value of *[tex \Large P_4(4)] from question 1.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>