Question

Consider two events A and B. It is known that P(A) = 0.1, P(B|A) = 0.1 and P(B|AC) = 0.4. Calculate the following. Enter your answers to at least four decimal places accuracy. (a) P(AC) = 3 (b) P(B) = 3 (c) P(ANB) = 3 (d) P(A|B) = 3

Answer 1

Given,

P(A) = 0.1, P(B/A) = 0.1, P(B/A^C ) = 0.4

a) We know that P(A^C ) = 1 - P(A)

$\therefore$P(A^c ) = 1 - 0.1 = 0.9

b) P(B/A) = P(A$\cap$B)/P(A)

0.1=P(A$\cap$B)/0.1

P(A$\cap$B) = 0.1*0.1 = 0.01 ......................... (eq 1)

So,

P(A/B) = P(A$\cap$B)/P(B)

P(A$\cap$B)= P(B)*P(A/B)

From eq 1

0.01 = P(B)*P(A/B) ........................ (eq 2)

We know that,

P(B/A^c ) = P(A^c$\cap$B)/P(A^c)

P(A^c$\cap$B) = P(A^c)*P(B/A^C)

P(A^c$\cap$B) = 0.9 * 0.4

P(A^c$\cap$B) = 0.36

P(A^c/B) = P(A^c$\cap$B)/P(B)

P(A^c/B) = 1- P(A/B) ......... (From theorem of conditional probability)

P(A^c$\cap$B) = P(B)*P(A^c/B)

P(A^c$\cap$B) = P(B)* (1 - P(A/B))

0.36 = P(B) - {P(B)*P(A/B)}

From eq 2.

0.36 = P(B) - 0.01

P(B) = 0.36 + 0.01

P(B) = 0.37

c) P(A$\cap$B) = 0.01

d) P(A/B) = P(A$\cap$B)/P(B)

P(A/B) = 0.01/0.37

P(A/B) = 0.0270