Question

an internet research company surveyed 85 online shoppers, each of whom made one purchase today. the company recorded the type of purchase each shopper made. here is a summary.
type of purchase number of shoppers
toys 40
office supplies 12
beauty supplies 20
food 13
three shoppers from the survey are selected at random, one at a time without replacement. what is the probability that the first shopper purchased office supplies and the other two did not?
do not round your intermediate computations. round your final answer to three decimal places.

Explanation:

Step1: Calculate probability of first - shopper buying office supplies

The total number of shoppers is $n = 85$, and the number of shoppers who bought office supplies is $12$. The probability that the first shopper purchased office supplies is $P_1=\frac{12}{85}$.

Step2: Calculate probability of second - shopper not buying office supplies

After the first shopper is selected, there are $n_1 = 84$ shoppers left. The number of shoppers who did not buy office supplies is $85 - 12=73$. So the probability that the second shopper did not purchase office supplies is $P_2=\frac{73}{84}$.

Step3: Calculate probability of third - shopper not buying office supplies

After the second shopper is selected, there are $n_2 = 83$ shoppers left. The probability that the third shopper did not purchase office supplies is $P_3=\frac{72}{83}$.

Step4: Calculate the overall probability

Since these are sequential and independent - like (in the non - replacement sense) events, the overall probability $P$ is the product of the individual probabilities. So $P = P_1\times P_2\times P_3=\frac{12}{85}\times\frac{73}{84}\times\frac{72}{83}$.
\[

$$\begin{align*} P&=\frac{12\times73\times72}{85\times84\times83}\\ &=\frac{12\times73\times72}{85\times7032}\\ &=\frac{12\times73\times72}{597720}\\ &=\frac{12\times5256}{597720}\\ &=\frac{63072}{597720}\approx 0.105 \end{align*}$$

\]

Answer:

$0.105$