In the first quarter of a year, a company’s records showed that 68% of its sales employees missed no work and 76% of its sales employees made their sales quota for the quarter. Of the sales employees who missed no work, 87% made their sales quota for the quarter. What is the probability that a randomly selected sales employee in the company missed no work and made their sales quota for the quarter
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
.68 * .87 = ?
To find the probability that a randomly selected sales employee in the company missed no work and made their sales quota for the quarter, you need to use conditional probability.
First, let's define the events:
A = Missed no work
B = Made sales quota
The information given in the question is as follows:
P(A) = 68% = 0.68 (probability of missing no work)
P(B) = 76% = 0.76 (probability of making sales quota)
P(B|A) = 87% = 0.87 (probability of making sales quota given that no work was missed)
To find the probability that a sales employee missed no work and made their sales quota, you can use the formula for conditional probability:
P(A and B) = P(A) * P(B|A)
Substituting the given values:
P(A and B) = 0.68 * 0.87
= 0.5916
Therefore, the probability that a randomly selected sales employee in the company missed no work and made their sales quota for the quarter is approximately 0.5916 or 59.16%.