Question

Use the formulas p(A" and " B)=p(A)+p(B)-p(A" or " B) and p(A" given " B)=p(A" and " B)/p(B) to answer the following question:Suppose on a particular day, the probability (among the entire population) of getting into a car accident is 0.04, the probability of being a texter-and-driver is 0.14, and p(car accident or being a texter-and-driver)=0.15. Find the probability a person was in a car accident given that they are a texter-and-driver. Is this higher or lower than the probability among the general population and why?Remember the formula for expected value is: EV = Outcome1 * Probability1 + Outcome 2 * Probability2… and so on.

Answer 1

We have the next given information:

Set X for the probability of getting into a car accident.

P(X) = 0.04

Set Y for the probability of being a texter-and-driver is 0.14

P(Y)=0.14

Then, the probability of being a texter-and-driver is 0.14, and p(car accident or being a texter-and-driver):

P(X or Y)= 0.15

Now, we need to find the probability a person was in a car accident given that they are a texter-and-driver.

This represents the conditional P(X/Y) = P(X and Y)/ P(Y)

Where :

(X and Y) = P(X)+P(Y)-P(X or Y)

(X and Y) = 0.004+0.14-0.15

(X and Y) =0.03

Replacing on the conditional equation:

P(X/Y) = P(X and Y)/ P(Y)

P(X/Y) = 0.03/ 0.4

P(X/Y) = 0.2143

Therefore, the probability a person was in a car accident given that they are a texter-and-driver is 0.2143.

The probability is higher than the probability among the general population.

P(X/Y) > P(X)