Discrete Random Variables: Question 4

Worked solution

Step 1: Identify n, p and r

$X\sim B(8,0.3)$, so $n=8$, $p=0.3$, $q=1-p=0.7$. We need $P(X=2)$, so $r=2$.

Step 2: Apply the binomial formula

$P(X=r)=\binom{n}{r}p^r q^{n-r}$ $P(X=2)=\binom{8}{2}(0.3)^2(0.7)^{6}$

Compute each part: $\binom{8}{2}=28, \qquad (0.3)^2=0.09, \qquad (0.7)^{6}\approx 0.117649$

Multiplying: $P(X=2)=28\times0.09\times0.117649\approx 0.2965$

So $P(X=2)\approx \boxed{0.296}$ (to 3 significant figures).

Check by a different grouping: $28\times0.09=2.52$, and $2.52\times0.117649\approx0.29648$, which rounds to the same $0.296$.

Why the other options are wrong

• B ($0.254$): this is actually $P(X=3)=\binom{8}{3}(0.3)^3(0.7)^5$, obtained by using the wrong value of $r$.
• C ($0.424$): this comes from using $(0.7)^5$ instead of $(0.7)^6$, an off-by-one error in the exponent $n-r=8-2=6$.
• D ($0.0106$): this is $(0.3)^2(0.7)^6$ with the binomial coefficient $\binom{8}{2}=28$ left out altogether.

Final answer

• $P(X=2)\approx \boxed{0.296}$, option A.