Question

(1 point) If P(En F) = 0.036, P(E|F) = 0.12, and P(F|E) = 0.4, then (a) P(E) = (b) P(F) = (c) P(EUF) = (d) Are the events E and F independent? Enter yes or no

Answer 1

By Bayes Theorem, P(A / B) = P(A $\cap$ B) / P(B)

_________________________________________

(a) P(F / E) = P(F $\cap$ E) / P(E)

Therefore P(E) = P(F $\cap$ E) / P(F / E) = 0.036 / 0.4 = 0.09

________________________________________

(b) P(E / F) = P(E $\cap$ F) / P(F)

Therefore P(F) = P(E $\cap$ F) / P(E / F) = 0.036 / 0.12 = 0.3

________________________________________

(c) P(E U F) = P(E) + P(F) - P(E $\cap$ F)

Therefore P(E U F) = 0.3 + 0.09 - 0.036 = 0.354

________________________________________

(d) For 2 events A and B to be independent, P(A $\cap$ B) = P(A) x P(B)

P(E $\cap$ F) = 0.036 and

P(E) x P(F) = 0.3 x 0.09 = 0.027

Since P(E $\cap$ F) $\neq$ P(E) x P(F), therefore,

NO the 2 events are not independent.

________________________________________

Answer 2

#6 A probability p is defined on the sample space Ω. Consider two events E and F defined by P(E) = 0.3 and P(F) = 0.4. Calculate P( E or F ) in each of the following cases: a. F occurs if E is occurred. b. E and F are independent. c. E and F are mutually exclusive events.