what's the probability that x=4 in the binomial experiment where n=9 and p=.3?

Question

anonymous · Answer

Consider that $x=4$ with $n=9$ implies four out of nine Bernoulli trials ended in success; the number of ways this could occur is given by $_9C_4=\dfrac{9!}{5!4!}=\frac{9\times8\times7\times6}{4\times3\times2}=9\times2\times7=126$. Given each trial is independent, and the probability for a trial to end in success is $p=0.3$ and failure $1-p=0.7$ the probability of any particular way of getting four out of nine trials is $p^4(1-p)^{9-4}=0.3^40.7^5$ and counting all the different possible ways gives a final probability of $_9C_4p^4(1-p)^{9-4}=126(0.3)^4(0.7)^5\approx0.172$ i.e. $17.2\%$.

anonymous · Answer

In general, for $X\sim\text{Binom}(n,p)$ we have $\mathbb{P}(X=x)=\,_nC_xp^x(1-p)^{n-x}$