A man speaks the truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even. The probability that it is actually even is

Concept:

E = the event a man reports that it is even

E1 = the event of an even number is picked

E2 = the event of an odd number is picked

To find : The probability that it is actually even i.e. P(E1|E)

P(E1|E) = \(\rm \dfrac {P(E|E_1).P(E_1)}{P(E|E_1).P(E_1)+P(E|E_2).P(E_2)}\)

Calculations:

Given, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even.

E₁ = the event of an even number is picked

E₂ = the event of an odd number is picked

⇒P(E1) = \(\dfrac 3 7\)

⇒P(E2) = \(\dfrac 4 7\)

E = the event a man reports that it is even

 A man speaks truth 2 out of 3 times.

⇒P(E|E₁) = \(\dfrac 2 3\)

⇒P(E|E2) = \(\dfrac 1 3\)

To find : The probability that it is actually even i.e. P(E₁|E)

P(E1|E) = \(\rm \dfrac {P(E|E_1).P(E_1)}{P(E|E_1).P(E_1)+P(E|E_2).P(E_2)}\)

⇒P(E1|E) = \(\rm \dfrac {(\dfrac 2 3).(\dfrac 3 7)}{(\dfrac 2 3).(\dfrac 3 7)+(\dfrac 1 3).(\dfrac 4 7)}\)

⇒P(E1|E) = \(\rm \dfrac 6 {10}\)

⇒⇒P(E1|E) = \(\rm \dfrac 3 5\)

Hence, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even. The probability that it is actually even is \(\rm \dfrac 3 5\)