By the end of 2022, the percentages of a population who has received V dose(s) of Covid-19 vaccine are shown in the table below, with no one receiving 5 or more doses of vaccine. Mr Lee is doing a survey on people's perspectives on the Covid-19 vaccine. [Continued] Assume now that these assumptions do in fact hold.

1. Formulate an equation using the fact that the sum of probabilities is 1.

Let $P(V=v)$ be the probability that a person receives $v$ doses.
$\sum P(V=v) = 1$
$0.07 + \frac{a}{100} + \frac{b}{100} + 0.69 + 0.09 = 1$
$\frac{a+b}{100} = 1 - 0.85 = 0.15$
$a + b = 15 \quad \text{--- (1)}$

2. Formulate a second equation using the given expectation.

$E(V) = \sum v \cdot P(V=v) = 2.72$
$0(0.07) + 1\left(\frac{a}{100}\right) + 2\left(\frac{b}{100}\right) + 3(0.69) + 4(0.09) = 2.72$
$\frac{a + 2b}{100} + 2.07 + 0.36 = 2.72$
$\frac{a + 2b}{100} = 2.72 - 2.43 = 0.29$
$a + 2b = 29 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $a$ and $b$.

Subtracting equation (1) from equation (2):
$(a + 2b) - (a + b) = 29 - 15$
$b = 14$

Substitute $b = 14$ into equation (1):
$a + 14 = 15 \implies a = 1$

Thus, $P(V=1) = 0.01$ and $P(V=2) = 0.14$.

4. Calculate the variance of $V$.

$E(V^2) = \sum v^2 \cdot P(V=v)$
$E(V^2) = 0^2(0.07) + 1^2(0.01) + 2^2(0.14) + 3^2(0.69) + 4^2(0.09)$
$E(V^2) = 0 + 0.01 + 0.56 + 6.21 + 1.44 = 8.22$

$\operatorname{Var}(V) = E(V^2) - [E(V)]^2$
$\operatorname{Var}(V) = 8.22 - (2.72)^2$
$\operatorname{Var}(V) = 8.22 - 7.3984 = 0.8216$

1. Identify the possible sum combinations for 3 people to have more than 10 doses in total.

Let $T$ be the total number of doses received by 3 people. Since the maximum number of doses per person is $4$, the only possible totals greater than $10$ are $11$ and $12$.

For $T = 12$, the only possible combination is $(4, 4, 4)$.
For $T = 11$, the possible combinations are $(4, 4, 3)$, $(4, 3, 4)$, and $(3, 4, 4)$.

2. Calculate the probability for each of the valid totals using the independent probabilities from the table.

From the table, $P(V=4) = 0.09$ and $P(V=3) = 0.69$.

$P(T=12) = P(V=4)^3 = (0.09)^3$
$P(T=11) = 3 \times P(V=4)^2 \times P(V=3) = 3 \times (0.09)^2 \times 0.69$

3. Add the probabilities of these mutually exclusive cases to find the final probability.

$P(T > 10) = P(T=12) + P(T=11)$
$P(T > 10) = (0.09)^3 + 3(0.09)^2(0.69)$
$P(T > 10) = 0.000729 + 0.016767 = 0.017496$

1. State the assumption of independence in the context of the problem.

Whether a walk-in opts for Moderna's bivalent vaccine is independent of the choice made by any other walk-in.

2. State the assumption of constant probability in the context of the problem.

The probability that any walk-in chooses Moderna's bivalent vaccine is a constant at $0.6$ (or $60\%$).

1. Identify the parameters of the binomial distribution and express the required probability interval in terms of inclusive discrete values.

Given that $n = 10$ and the probability of a walk-in opting for Moderna's bivalent vaccine is $60\%$, we have:
$M \sim \mathrm{B}(10, 0.6)$
Since $M$ represents a discrete number of people, the inequality $4 \le M < 8$ includes the integers $4, 5, 6,$ and $7$.
$\mathrm{P}(4 \le M < 8) = \mathrm{P}(4 \le M \le 7)$

2. Evaluate the probability using the binomial cumulative distribution function.

$\mathrm{P}(4 \le M \le 7) = \mathrm{P}(M \le 7) - \mathrm{P}(M \le 3)$
Using a graphing calculator:
$\mathrm{P}(4 \le M \le 7) \approx 0.832710 - 0.054762$
$= 0.777948...$
$\approx 0.778 \text{ (to 3 s.f.)}$

1. Formulate the inequality for the given probability condition.

$M \sim \mathrm{B}(n, 0.6)$
We require $\mathrm{P}(M \ge 15) \ge 0.9$.
$1 - \mathrm{P}(M \le 14) \ge 0.9$
$\mathrm{P}(M \le 14) \le 0.1$

2. Use a graphing calculator to evaluate the probabilities for different values of $n$.

Using a graphing calculator, we check the values of $\mathrm{P}(M \le 14)$ for consecutive integer values of $n$:
When $n = 29$, $\mathrm{P}(M \le 14) \approx 0.1339 > 0.1$
When $n = 30$, $\mathrm{P}(M \le 14) \approx 0.0940 \le 0.1$

3. State the least number of people to be surveyed.

Since the condition is first met when $n = 30$, the least number of people Mr Lee needs to survey is $30$.