Question:

Given that $\displaystyle Y \sim B(7,\frac{2}{3})$, calculate $\displaystyle P(Y=4$).

Attempt:

$\displaystyle n = 7, p = \frac{2}{3}, q = 1-\frac{2}{3} = \frac{1}{3}$

$\displaystyle P(Y=y) = ^nC_y\times p^n\times q^{n-y}$

$\displaystyle P(Y=4) = ^7C_4\times \frac{2}{3}^4\times \frac{1}{3}^{7-4}$

$\displaystyle

P(Y=4) = 0.032$

Where did I go wrong?