Answer to Question #125372 in Statistics and Probability for @

Question

Traffic police checked the CNICs and driving licenses of all the people driving any vehicle on a particular road on a given day. 85% of the drivers were carrying valid CNICs and 75% of the drivers were carrying valid driving licenses. 65% of the drivers were carrying both valid CNICs and driving licenses. What percent of those who were carrying a valid CNIC were also carrying a valid driving license? What percent of those who were carrying a valid driving license were also carrying a valid CNIC?

Answer 1

Let, A = the event that a randomly selected person was carrying valid CNICs

and B = the event that a randomly selected person was carrying valid driving license

$\therefore$ A $\cap$ B = the event that a randomly selected person was carrying both valid CNICs and driving license

Then we are given,

P(A) = 85% = 0.85, P(B) = 75% = 0.75 and P(A $\cap$ B) = 65% = 0.65

$\therefore$ The probability that a person who was carrying a valid CNIC was also carrying a valid driving license

= The probability that a person was carrying a valid CNIC given that he was carrying a valid driving license

= P(A | B)

= $\frac{P(A\cap B)}{P(B)}$ = $\frac{0.65}{0.75}=0.87$

And the probability that a person who was carrying a valid driving license was also carrying a valid CNIC

= The probability that a person was carrying a valid driving license given that he was carrying a valid CNIC

= P(B | A)

= $\frac{P(A\cap B)}{P(A)}$

= $\frac{0.65}{0.85}=0.77$

Answer: 87% of those who were carrying a valid CNIC were also carrying a valid driving license. 77% of those who were carrying a valid driving license were also carrying a valid CNIC.