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In this article, we will discuss what is conditional probability and how to calculate it.
What is a Conditional Probability?
The conditional probability of event A is the probability that the event will occur given the information that event B has already occurred. We can denote this probability as P(A|B). It is read as the probability of A given B.
There are two types of events: Independent and dependent events.
Independent events are not dependent on each other. An example of an independent event includes rolling a die. Each time a die is rolled the probability of getting any number remains $\frac {1}{6}$.
The dependent event is affected by a posterior event. Suppose that there are 8 balls on the floor. 2 of the 8 balls are red and the remaining are blue. If you pick any ball randomly from the floor, then the probability of picking up a red ball is $\frac {2}{8} = \frac {1}{4}$. If another ball is picked up, then the probability of getting a red ball will be $\frac {1}{7}$.
You can see that the probability has changed because after picking up one red ball from the floor, the total number of balls left was 7, and 1 red ball was left. It means that the second event is dependent on the first event.
If the events A and B are independent, i.e., the occurrence of the event B does not affect the probability of event A, then the conditional probability of event A given event B is simply the probability of event A, which is denoted by P(A).
Formula for Calculating the Conditional Probability
The conditional probability of an event A, given that event B has already occurred can be computed using the following formula:
$P(A|B) = \frac {P (A \cap B)} {P(B)}$
Here:
P(A|B) represents the probability that an event A will occur, given that event B has already occurred
$P (A \cap B)$ reflects the probability that an event A and B will occur
P(B) represents the probability of an event B
We can also reverse the above formula if we have to find the probability of an event B given that event A has already occurred. In this case, the formula will be:
$P(B|A) = \frac {P (A \cap B)} {P(A)}$
P(A|B) represents the probability that an event B will occur, given that event A has already occurred
$P (A \cap B)$ reflects the probability that an event A and B will occur
P(A) represents the probability of an event A
In the next section, we will solve some examples related to conditional probability.
Example 1
What is the probability of getting 4 on the die if it is already known that a result is an even number?
Solution
Total number of outcomes = 6
Probability of getting 4 = P(A) = $\frac {1}{6}$
Total number of even outcomes = (2, 4, 6) = 3
Probability of getting an even number = P(B)= $\frac {3}{6} = \frac {1}{2}$
Probability of getting 4 on a die, given that the result is even = $P(A|B) = \frac {P (A \cap B)} {P(B)}$
= $\frac{\frac{1}{6}} {\frac{1}{2}} = \frac{1}{3}$
Example 2
100 students participated in a survey conducted at a school. Out of 100 students, 30 said that they like mathematics, 25 said that they like science, and 35 said that they like mathematics and science. If a student chosen randomly likes mathematics, what is the probability that he/she also likes science?
Solution
The probability that a randomly chosen student likes mathematics = P(A) = 0.40
Probability that a randomly chosen student likes science and mathematics = $P(A \cap B)$ = 0.35
Probability that a randomly chosen student likes mathematics, given that he/she also likes science = P(B|A) = ?
We will find the unknown probability by substituting the values in the formula below:
$P(B|A) = \frac {P (A \cap B)} {P(A)}$
$P(B|A) = \frac {0.35} {0.40}$
= $0.875$
Hence, the probability that a randomly selected student likes mathematics, given that he/she already likes science is 0.875 or 87.5%.
Example 3
The probability that it is Sunday and there is traffic on the road is 0.09. The probability that it is Sunday is 0.14. What is the probability that there is traffic on the road, given that it is Sunday?
Solution
The probability that it is Sunday= P(A) = 0.14
The probability that it is Sunday and there is a traffic on the road= $P(A \cap B)$ = 0.09
The probability that there is traffic on the road given that it is Sunday = P(B|A) = ?
We will use the following formula to calculate the probability:
$P(B|A) = \frac {P (A \cap B)} {P(A)}$
Substitute the values in the above formula:
$P(B|A) = \frac {0.09} {0.14} = 0.64$
Hence, there is a 0.64 probability or 64% chance that there is traffic on the road and it is Sunday.
Example 4
At a business school, the probability that a student takes all the courses in one semester and does a job is 0.045. The probability that the student takes all the courses in the semester is 0.75. What is the probability that the student does a job, given that he/she has taken all the courses in that semester?
Solution
The probability of taking all the courses in a semester and doing a job = $P(A \cap B) = 0.045
The probability that a student takes all the courses in the semester = P(A) = 0.75
The probability of student doing a job, given that he/she is taking all the course = P(B|A) = ?
We will use the following formula to calculate the probability:
$P(B|A) = \frac {P (A \cap B)} {P(A)}$
= $\frac{0.045}{0.75}$
= $0.06$
Hence, the probability that a student is doing a job, given that he/she has taken all the courses is 0.06 or 6%.
Example 5
In an area, 25% of the population own a car and a bike. 50% of the population own a car only. What is the probability that a randomly chosen person owns a bike, given that he already owns a car?
Solution
The probability that a randomly chosen person owns a car and a bike = $P(A \cap B)$ = 0.25
The probability that a randomly chosen person owns a car only = P(A) = 0.5
The probability that a randomly chosen person owns a bike, given that he/she already owns a car = P(B|A) = ?
We will use the following formula to compute this probability:
$P(B|A) = \frac {P (A \cap B)} {P(A)}$
Substitute the values in the above formula to get the probability:
$P(B|A) = \frac {0.25} {0.50}$
$P(B|A) = 0.50$
Hence, there is a 0.50 probability or 50% chance that a randomly chosen person owns a bike given that he already owns a car.
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