The Binomial Distribution

Let $X \sim \mathrm{Bi}(4, p)$ be a binomial random variable. Show that $\Pr(X \geq 3) = g(p)$ for all $p \in [0, 1]$, where $g(p) = 4p^3 - 3p^4$.

1. 1

   'At least three' out of four means $X = 3$ or $X = 4$. Add the two binomial probabilities, taking care to include both terms.

   $$\Pr(X \geq 3) = \binom{4}{3}p^3(1-p) + \binom{4}{4}p^4$$
2. 2

   Evaluate the coefficients $\binom{4}{3} = 4$ and $\binom{4}{4} = 1$, then expand and collect like terms.

   $$4p^3(1-p) + p^4 = 4p^3 - 4p^4 + p^4 = 4p^3 - 3p^4 = g(p)$$
3. 3

   As a 'show that', full working is required. The examiner report notes some students computed $\Pr(X \leq 3)$ or omitted the $X = 4$ term entirely.

   $$\Pr(X \geq 3) = 4p^3 - 3p^4 = g(p)$$

VCAA 2025 Mathematical Methods Exam 2, Section B Q1f

Let $X \sim \mathrm{Bi}\left(4, \dfrac{9}{10}\right)$. Find $\Pr(X < 2)$.

1. 1

   $\Pr(X < 2)$ means $X = 0$ or $X = 1$. Write each term with the binomial formula, keeping the success power and failure power matched.

   $$\Pr(X=0) = \left(\tfrac{1}{10}\right)^4, \quad \Pr(X=1) = \binom{4}{1}\left(\tfrac{9}{10}\right)\left(\tfrac{1}{10}\right)^3$$
2. 2

   Evaluate each term over the common denominator $10^4 = 10000$.

   $$\Pr(X=0) = \frac{1}{10000}, \quad \Pr(X=1) = \frac{36}{10000}$$
3. 3

   Add the two terms. The examiner report notes some students mishandled the decimal or fraction expansion (an extra or missing zero) or gave only the $\Pr(X=1)$ term.

   $$\Pr(X < 2) = \frac{1}{10000} + \frac{36}{10000} = \frac{37}{10000}$$

VCAA 2024 Mathematical Methods Exam 1, Q4b

A basketball player has a $0.85$ chance of sinking each free throw, and each throw is independent. She takes $8$ throws. Find the probability that she sinks exactly $6$ of them, correct to four decimal places.

1. 1

   This is binomial. There are $n=8$ fixed trials, two outcomes each (sink or miss), a constant success chance $p=0.85$, and the throws are independent. We want exactly $k=6$ successes.

   $$X \sim \text{Bi}(8,\ 0.85)$$
2. 2

   Write the binomial formula. The $\binom{8}{6}$ counts the different orders in which the $6$ successes could land among the $8$ throws.

   $$\Pr(X=6) = \binom{8}{6}(0.85)^6(0.15)^2$$
3. 3

   Evaluate. Keep the success power and the failure power matched to $6$ and $2$, never the other way around.

   $$\Pr(X=6) = 28 \times (0.85)^6 \times (0.15)^2$$

A factory line produces phone screens, and each screen independently has a $0.2$ chance of a tiny scratch. A box holds $12$ screens. Find the mean, the variance and the standard deviation of the number of scratched screens in a box.

1. 1

   Set it up as binomial with $n=12$ and $p=0.2$.

   $$X \sim \text{Bi}(12,\ 0.2)$$
2. 2

   The mean of a binomial is $np$. On average, a box has this many scratched screens.

   $$E(X) = np = 12 \times 0.2 = 2.4$$
3. 3

   The variance is $np(1-p)$. This measures spread, not a typical count.

   $$\text{Var}(X) = np(1-p) = 12 \times 0.2 \times 0.8 = 1.92$$
4. 4

   The standard deviation is the square root of the variance. This is the step students most often forget.

   $$sd(X) = \sqrt{1.92}$$

A seed company says each seed germinates independently with probability $0.97$. A gardener plants $5$ seeds. Find the probability that at least one seed fails to germinate, correct to four decimal places.

1. 1

   Let $X$ be the number that fail. A failure has probability $1-0.97=0.03$, so $X \sim \text{Bi}(5,\ 0.03)$. 'At least one' means $X \geq 1$, which is everything except zero.

   $$\Pr(X \geq 1) = 1 - \Pr(X=0)$$
2. 2

   Find $\Pr(X=0)$ first. With zero failures, all five must germinate.

   $$\Pr(X=0) = \binom{5}{0}(0.03)^0(0.97)^5 = (0.97)^5$$
3. 3

   Subtract from one. Going through the complement is far quicker than adding up the cases for $X=1,2,3,4,5$.

   $$\Pr(X \geq 1) = 1 - (0.97)^5$$