Kiran surveyed a random sample of students in her school district, and she found these statistics:
P{rides bike to school})=0.1
P{has crossing guard})=0.48
P{rides bike and crossing guard})=0.12

Find the probability that a student rides a bike to school, given that the student's school has a crossing guard.

Probability that event A or B occurs = Probability that event A occurs + Probability that event B occurs - Probability that both the event A and B occur simultaneously.

This can be written symbolically as:

[tex]P( A \cup B ) = P(A) + P(B) - P(A \cap B)[/tex]

What is the chain rule in probability for two events?

For two events A and B:

The chain rule states that the probability that A and B both occur is given by:

[tex]P(A \cap B) = P(A)P(B|A) = P(B)P(A|B)[/tex]

You can remember this fact by using a trick that when A is on left, it is behind B in next term P(A)P(B|A)

And when B is on left, then it is behind A in next term P(B)P(A|B)

P(A|B) = Probability that A will happen given B has already happened.

P(B|A) = Probability that B will happen given A has already happened.

Using those above stated rules, we have

Let the event A be that a student rides bike to school.

Let the event B be that the student's school has a crossing guard

Then we are given that:

P(A) = P{rides bike to school})=0.1

P(B) = P{has crossing guard})=0.48

[tex]P(A \cap B)[/tex] = P{rides bike and crossing guard})=0.12

We have to find P(a student rides a bike to school, given that the student's school has a crossing guard) = P(A | B)

From the chain rule of probability, we have:
[tex]P(A \cap B) = P(B)P(A|B)[/tex]

Thus,

[tex]0.12 = 0.48 \times P(A|B)\\\\P(A|B) = \dfrac{0.12}{0.48} = 0.25[/tex]

Thus,

The probability that a student rides a bike to school, given that the student's school has a crossing guard is given by 0.25

Learn more about conditional probability here:

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