Discrete Random Variables: Question 3

Worked solution

Part (a): Conditions for a binomial model

For $X$ to be modelled by a binomial distribution $B(n,p)$, the trials must satisfy:

1. There is a fixed number of trials, here $n=10$ bulbs are sampled.
2. Each trial (bulb) has exactly two possible outcomes, “defective” or “not defective”.
3. The probability of a bulb being defective is constant for every bulb, here $p=0.08$.
4. The bulbs are independent of one another (one bulb being defective does not affect whether another is).

Any two of these conditions is sufficient for the marks, e.g. “the probability that a bulb is defective is constant at $0.08$ for every bulb in the sample, and whether or not one bulb is defective is independent of any other bulb.”

Since these hold, $X\sim B(10,0.08)$.

Part (b): Finding P(X = 2)

Using $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$ with $n=10$, $p=0.08$, $r=2$: $P(X=2)=\binom{10}{2}(0.08)^2(0.92)^{8}$

Compute each piece: $\binom{10}{2}=45, \qquad (0.08)^2=0.0064, \qquad (0.92)^{8}\approx 0.51322$

Multiplying: $P(X=2)=45\times0.0064\times0.51322\approx 0.1478$

So $P(X=2)\approx \boxed{0.148}$ (to 3 significant figures).

Check by an alternative grouping: $45\times0.0064=0.288$, and $0.288\times0.51322\approx0.14781$, which rounds to the same $0.148$.

Part (c): Finding P(X ≥ 2)

“At least $2$” is the complement of ”$0$ or $1$”, so $P(X\geq 2)=1-P(X=0)-P(X=1)$

Find $P(X=0)$: $P(X=0)=\binom{10}{0}(0.08)^0(0.92)^{10}=(0.92)^{10}\approx 0.4344$

Find $P(X=1)$: $P(X=1)=\binom{10}{1}(0.08)^1(0.92)^{9}=10\times0.08\times(0.92)^{9}\approx 10\times0.08\times0.47216\approx 0.3777$

So $P(X\geq 2)=1-0.4344-0.3777=0.1879$

$P(X\geq 2)\approx \boxed{0.188}$ (to 3 significant figures).

Check by direct summation: adding $P(X=r)$ for $r=2,3,\ldots,10$ term by term gives the same total, $0.1879$ (to 4 d.p.), which confirms the complement calculation.

Part (d): Finding E(X)

For a binomial distribution, $E(X)=np$: $E(X)=10\times0.08=0.8$

Final answers

• (a) Any two valid conditions, e.g. constant probability $p=0.08$ and independence between bulbs.
• (b) $P(X=2)\approx\boxed{0.148}$
• (c) $P(X\geq2)\approx\boxed{0.188}$
• (d) $E(X)=\boxed{0.8}$