# Math Help - Binomial Distribution...

1. ## Binomial Distribution...

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

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

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

$P(Y=4) = ^7C_4\times \frac{2}{3}^4\times \frac{1}{3}^{7-4}$
$
P(Y=4) = 0.032$

Where did I go wrong?

2. Try using brackets on your fractions.

$P(Y=4) = ^7C_4\times (\frac{2}{3})^4\times (\frac{1}{3})^{7-4}=0.256$

Is that the answer you expected?

3. Originally Posted by a tutor
Try using brackets on your fractions.

$P(Y=4) = ^7C_4\times (\frac{2}{3})^4\times (\frac{1}{3})^{7-4}=0.256$

Is that the answer you expected?

$P(Y=4) = ^7C_4\times \left( \frac{2}{3} \right)^4 \times \left( \frac{1}{3} \right)^{7-4}=0.256$