Question

If $X \sim B(4, p)$ and $P(X = 0) = \frac{16}{81}$, then $P(X = 4) =$

• A. $\frac{1}{81}$
• B. $\frac{1}{16}$
• C. $\frac{1}{8}$
• D. $\frac{1}{27}$

Hint:

Notice the symmetry of the exponents in binomial boundaries! For $X = 0$, the value is simply $q^n$, and for $X = n$, the value is simply $p^n$. Since $q^4 = (2/3)^4$, its complement must be $p = 1/3$. Raising that to the fourth power gives $\frac{1}{81}$ in one quick step!

Answer 1

The Correct Option is A

Step 1: Write the no-success probability.
With $n=4$, $P(X=0)=q^4$ where $q=1-p$. We are told this equals $\frac{16}{81}$.

Step 2: Find $q$.
$q^4=\frac{16}{81}$, so $q=\sqrt[4]{\frac{16}{81}}=\frac{2}{3}$.

Step 3: Find $p$.
$p=1-q=1-\frac{2}{3}=\frac{1}{3}$.

Step 4: Work out $P(X=4)$.
$P(X=4)=p^4=\left(\frac{1}{3}\right)^4=\frac{1}{81}$. \[ \boxed{\frac{1}{81}} \]