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.
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
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