Question

A firm has the following rules:
When a worker comes late there is ¼ chance that he is caught.
First time he is given a warning.
Second time he is dismissed.
What is the probability that a worker is late three times is not dismissed?

Answer 1

Answer on Question #38851, Math, Statistics and Probability

A firm has the following rules: When a worker comes late there is $\frac{1}{4}$ chance that he is caught. First time he is given a warning. Second time he is dismissed. What is the probability that a worker is late three times is not dismissed?

Solution

For this task we will use the Bernoulli trials formula.

Let the event when worker is caught be defined as success and when he isn't caught as a failure. The probability of success is $\frac{1}{4}$ . Thus the probability of failure is $\frac{1}{4}$ . We need to find the following probability

$$P (k \leq 1) = P (0) + P (1) = \binom {0} {3} \left(\frac {1}{4}\right) ^ {0} \left(\frac {3}{4}\right) ^ {3} + \binom {1} {3} \left(\frac {1}{4}\right) ^ {1} \left(\frac {3}{4}\right) ^ {2} = \frac {2 7}{6 4} + \frac {2 7}{6 4} = \frac {5 4}{6 4}.$$

Answer: $\frac{54}{64}$ .