Question:
You buy a certain type of lottery ticket once a week for 4 weeks. What's the probability you win a cash prize exactly twice?
My attempt:
P(Success) = $1/4 $
P(Not Successful) =$ 3/4$
$$4 \ _nC^r \ 2 \times \bigg(\frac{1}{4}\bigg)^2 \times \bigg(\frac{3}{4}\bigg)^2 = 0.2109375$$
Is my answer correct?
Your reasoning is indeed correct. In order to check your calculation I highly recommend Wolfram Alpha :
Probability that X=2 if X has a binomial distribution with n=4 and p=.25
(0.210937).
Let $X\sim \text{Binom}\left(n=4,p=\frac{1}{4}\right)$, a random variable having a binomial distribution.
Then its probability mass function is $$ p(x) = b(x;n,p)= b(x;4,0.25) = \begin{cases} \displaystyle{\binom{4}{x}}(0.25)^x (0.75)^{4-x} &\mbox{ if } x = 0,1,2,3, 4, \\ \hspace{16mm}0 &\mbox{ otherwise}. \end{cases} $$ Then the probability of winning a cash prize exactly twice is $$ P(X=2)= p(2) = \displaystyle{\binom{4}{2}}(0.25)^2 (0.75)^{4-2} \approx 0.210938. $$
dbinom. With appropriate parameters it becomes:dbinom(2, 4, 1/4), which returns 0.2109375. So Yes, correct. // Here is how you can make ${4 \choose 2}$ in MathJax:${4 \choose 2}$– BruceET Oct 16 '17 at 02:27