Question

Suppose that your boss must choose three employees in your office to attend a conference in Jamaica. Because all 17 of you want to go, he decides that the only fair way is to draw names out of a hat. What is the probability that you, Anna, and Kevin are chosen? Enter a fraction or round your answer to 4 decimal places, if necessary.

Answer 1

Answer:

$P(You\ n\ Anna\ n\ Kevin) = \frac{1}{4913}$

Step-by-step explanation:

Given

$Total\ Employees = 17$

$To\ Select = 3$

Required

Determine the probability of selecting You, Anna and Kevin

Assuming the selection process is fair;

Each employee has a probability of;

$Probability = \frac{1}{17}$

So:

$P(You\ n\ Anna\ n\ Kevin) = P(You) * P(Anna) * P(Kevin)$

$P(You\ n\ Anna\ n\ Kevin) = \frac{1}{17} * \frac{1}{17} * \frac{1}{17}$

$P(You\ n\ Anna\ n\ Kevin) = \frac{1}{4913}$