A local bottling company has determined the number of machine breakdowns per month and their respective probabilities as shown below.




Number of Breakdowns
Probability

0
0.12

1
0.38

2
0.25

3
0.18

4
0.07






The probability of at least 3 breakdowns in a month is ____ .


A0.07
B0.25
C0.50
D0.75

Verified Answer :

  • The condition "at least 3 breakdowns" includes the mutually exclusive outcomes of exactly 3 breakdowns and exactly 4 breakdowns.
  • The individual probabilities are \(P(3) = 0.18\) and \(P(4) = 0.07\).
  • Summing these values gives the total probability: \(0.18 + 0.07 = 0.25\).
  • This matches option b. The final answer is \(\boxed{0.25}\).

Explanation

  1. Understand the problem and data analysis

The problem provides a probability distribution for the number of machine breakdowns per month at a bottling company. The possible outcomes (number of breakdowns) are \(0, 1, 2, 3, \text{ and } 4\), with the associated probabilities given in the table. We need to determine the probability of having "at least 3" breakdowns in a month.

  1. Identify the relevant events and apply the sum rule

The term "at least 3" means that the number of breakdowns can be \(3\) or any value greater than \(3\). Looking at our data, this includes the cases where the number of breakdowns is \(X = 3\) and \(X = 4\). Because these events are mutually exclusive (the machine cannot have both exactly 3 and exactly 4 breakdowns in the same month), we can find the total probability by summing their individual probabilities.

  1. Calculate the total probability

The probabilities for \(X=3\) and \(X=4\) are:

  • \(P(X = 3) = 0.18\)
  • \(P(X = 4) = 0.07\)

Now, we add these probabilities together:
\[P(X \geq 3) = P(X = 3) + P(X = 4)\]
\[P(X \geq 3) = 0.18 + 0.07 = 0.25\]

  1. Conclusion and final answer selection

Comparing our result, \(0.25\), to the given options:

  • a. \(0.07\)
  • b. \(0.25\)
  • c. \(0.50\)
  • d. \(0.75\)

The result matches option b. Thus, the probability of at least 3 breakdowns is \(\boxed{0.25}\).

Examples

This type of analysis is crucial in industrial management for predictive maintenance. If a manager knows the probability of having 3 or more breakdowns is 25%, they can allocate enough budget and spare parts for the month to ensure operations continue smoothly, effectively managing risks in a production line. This is a practical way to use probability distributions to make data-driven business decisions.