From a box containing 20 balls, 14 of them black and 6 white, we are drawing four balls with replacements. What is the probability, that black ball will be drawn twice?
is my solution right?
$n = 4, k = 2, p = 7/10$
using the formula:
i got $P(X = 2) = 0,2646$
Yes, it is correct, since replacement is allowed, each draw can be viewed as iid Bernoulli trial with probability of success $0.7$. The number of black balls out of $4$ trials is then $Bin(4, 0.7)$.
$$\binom{4}{2}(0.7)^2(0.3)^2=6(0.21)^2=6\left(\frac{21}{100} \right)^2=\frac{2646}{10000}$$