Question 874099
probability of success = .7 = p
probability of failure = 1 - .7 = .3 = q
n = 15
x = number of successes.


p(x) = nCx * p^x * q^(n-x)


replace values for variables in this formula to get:


p(5) = 15C5 * .7^5 * .3^10


15C5 = 3003


p(5) = 3003 * .16807 * .0000059049


p(5) = .002980287


that's your answer.


the probabilities for 1,2,3,4,5,6,..... are in the attached table.


the sum of the probabilities should be equal to 1 or something went wrong.


this table is good since the probabilities do sum up to be equal to 1.


<pre>
n	x	p	q	p^x	q^(n-x)	        nCx	p(x)
15	0	0.7	0.3	1.00	0.000000014349	1	0.000000014349
15	1	0.7	0.3	0.70	0.000000047830	15	0.000000502212
15	2	0.7	0.3	0.49	0.000000159432	105	0.000008202792
15	3	0.7	0.3	0.34	0.000000531441	455	0.000082939340
15	4	0.7	0.3	0.24	0.000001771470	1365	0.000580575378
15	5	0.7	0.3	0.17	0.000005904900	3003	0.002980286939
15	6	0.7	0.3	0.12	0.000019683000	5005	0.011590004761
15	7	0.7	0.3	0.08	0.000065610000	6435	0.034770014284
15	8	0.7	0.3	0.06	0.000218700000	6435	0.081130033329
15	9	0.7	0.3	0.04	0.000729000000	5005	0.147235986413
15	10	0.7	0.3	0.03	0.002430000000	3003	0.206130380978
15	11	0.7	0.3	0.02	0.008100000000	1365	0.218623131340
15	12	0.7	0.3	0.01	0.027000000000	455	0.170040213264
15	13	0.7	0.3	0.01	0.090000000000	105	0.091560114835
15	14	0.7	0.3	0.01	0.300000000000	15	0.030520038278
15	15	0.7	0.3	0.00	1.000000000000	1	0.004747561510
							
					total probability >>>	1.000000000000

</pre>


15C5 comes from the combination formula of nCx = n! / (x! * (n-1)!)


n! is equal to 15!


x! is equal to 5!


(n-x)! is equal to (15 - 5)! = 10!


nCx formula becomes 15! / (5! * 10!) = 3003