Solution

This situation can be described by a Binomial Distribution since we have:

• a fixed number ($9$) of trials (workers).
• each trial has two possible outcomes, “success” (the worker is absent) or “failure” (the worker is not absent).
• the probability of success (0.1) is constant.
• the outcome of each trial is independent of the outcome of all other trials.

Let $X$ denote the number of absent workers. Using the above formula, the probability that $2$ out of $9$ randomly chosen workers will be absent is:

\begin{align} \mathrm{P}(X =2) &= {}^9\mathrm{C}_2 \times 0.1^2 \times(1-0.1)^{9-2}\\ &=36 \times 0.1^2 \times(0.9)^{7}\\ &=0.172 \text{ (to 3 d.p.).}\\ \end{align}

See Example 2, showing how to calculate binomial probabilities if you are unsure about how the above probabilities were obtained.

The rest of the solution to this question can be found here.

Solution

Let $Y$ denote the number of customers who buy something. Here $n = 10$ and $p = 0.4$. So we have $Y \sim \mathrm{Bin}(10,0.4)$

(A)

To calculate the probability that exactly one customer buys something we use the formula above in the first orange box for a Binomial probability with $n=10, p = 0.4 \text{ and } r = 1$: \begin{align} \mathrm{P}(Y = 1) &= {}^{10}\mathrm{C}_1 \times 0.4^{1} \times(1 - 0.4)^{10-1}\\ &={}^{10}\mathrm{C}_1 \times0.4\times0.6^{9}\\ &=0.040\text{ ( 3 d.p.)}.\\ \end{align}

(B)

Using the same formula as in part (A) we have: \begin{align} \mathrm{P}(Y = 4) &= {}^{10}\mathrm{C}_4 \times 0.4^{4} \times(1 - 0.4)^{10-4}\\ &={}^{10}\mathrm{C}_4 \times0.4^{4}\times0.6^{6}\\ &=0.251 \text{ (3 d.p.)}.\\ \end{align}

(C)

Here we are required to calculate the cumulative probability $\mathrm{P}(Y \leq 2)$. Summing the probabilities $\mathrm{P}(Y=j)$ for $j=0, 1, 2$, we have:

\begin{align} \mathrm{P}(Y \leq 2) &= \mathrm{P}(Y=0) \; + \; \mathrm{P}(Y=1) + \; \mathrm{P}(Y=2)\\ &= \big({}^{10}\mathrm{C}_0 \times 0.4^{0} \times(1 - 0.4)^{10-0}\big) + \big({}^{10}\mathrm{C}_1 \times 0.4^{1} \times(1 - 0.4)^{10-1}\big) + \big({}^{10}\mathrm{C}_2 \times 0.4^{2} \times(1 - 0.4)^{10-2}\big)\\ &= \big({}^{10}\mathrm{C}_0 \times0.4^{0}\times0.6^{10}\big) + \big({}^{10}\mathrm{C}_1 \times0.4^{1}\times0.6^{9}\big)+ \big({}^{10}\mathrm{C}_2 \times0.4^{2}\times0.6^{8}\big)\\ &=0.0060+0.0403+0.1209\\ &=0.167\text{ (3 d.p.)}.\\ \end{align}

Calculating the above probability, we needed to add up only three binomial probabilities. However, what if we had to calculate $\mathrm{P}(Y\leq7)$? We would have to calculate eight different probabilities and then add them up! Fortunately, there are published tables of the cumulative probabilities of the binomial distribution available that we can use for this purpose.

Alternatively we could have used the fact that, since $n=10$ for this distribution (so $X \leq 10$) and the sum of the probabilities must be 1,

\begin{align} \mathrm{P}(Y\leq7)&=1- \mathrm{P}(Y\gt7)\\ &= 1-(\mathrm{P}(Y=8) + \mathrm{P}(Y=9) + \mathrm{P}(Y=10))\\ \end{align}

(D)

Here we wish to calculate $\mathrm{P}(Y\geq2)$.

Rather than calculating $\mathrm{P}(Y=2)+\mathrm{P}(Y=3)+...+\mathrm{P}(Y=10)$, we can instead use the fact that $\mathrm{P}(Y\geq2)=1-\mathrm{P}(Y\lt2)$. We then only need to calculate $\mathrm{P}(Y\leq1)$ and subtract this from $1$ to obtain the probability $\mathrm{P}(Y\geq2)$.

\begin{align} \mathrm{P}(Y\geq2) &= 1-\mathrm{P}(Y\leq1)\\ &= 1 - \mathrm{P}(Y=1) - \mathrm{P}(Y=0)\\ &= 1 - \big({}^{10}\mathrm{C}_1 \times 0.4^{1} \times(1 - 0.4)^{10-1}\big) - \big({}^{10}\mathrm{C}_0 \times 0.4^{0} \times(1 - 0.4)^{10-0}\big)\\ &= 1 - 0.0403 - 0.0060\\ &= 0.954 \text{ (3 d.p.)}.\\ \end{align}

(E)

Using the expectation formula, we would expect $10 \times 0.4 = 4$ customers to buy something.

Note that the variance is $10 \times 0.4 \times(1-0.4) = 2.4$.