+33661 Do you mean the probability of exactly 1 success in 4 trials given a probability of 0.1 of an individual success? If so, it is normal to put the 1 before the 4.
prob(k;n,p) = nCr(n,k)*pk(1-p)n-k n = number of trials (4); k = number of successes (1); p = probability of individual success (0.1).
$${\mathtt{prob}} = {\left({\frac{{\mathtt{4}}{!}}{{\mathtt{1}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.1}}}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{0.1}}\right)}^{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{1}}\right)} \Rightarrow {\mathtt{prob}} = {\mathtt{0.291\: \!6}}$$
+33661 Do you mean the probability of exactly 1 success in 4 trials given a probability of 0.1 of an individual success? If so, it is normal to put the 1 before the 4.
prob(k;n,p) = nCr(n,k)*pk(1-p)n-k n = number of trials (4); k = number of successes (1); p = probability of individual success (0.1).
$${\mathtt{prob}} = {\left({\frac{{\mathtt{4}}{!}}{{\mathtt{1}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.1}}}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{0.1}}\right)}^{\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{1}}\right)} \Rightarrow {\mathtt{prob}} = {\mathtt{0.291\: \!6}}$$
