Question

Exercise 2 In a state, the probability that a person speaks French is 0.55. It is known that 15% from persons who speak French and 25% from persons who don't speak French are teachers. A person is selected at random. (1) Determine the probability that the selected person is a teacher (2) Suppose that the selected person is a teacher. What is the probability that she speaks French? (3) Determine the probability that the selected person is a teacher or speaks French

Answer 1

PIA) = 0.62 PCB) = 0 49 PLANB) = 0.38 PLAUB) = PLA) + P (6) - PLANB) AMB Ano) 0.627 0.49 - 0-38 0.73 AnB ** CD PLANB') = ? LLLPPPPPPP Since P LA OB' = IP (AUB). = 10:13 = 0.27 (0) PCA U B') - P(A) + P (BP) - PLANB!) = [-P(A)] + [ 1-PIB)} - PLANB') = 2-PIA) - P (B)- PIANB'). = 2 - 0.62 - 0.49 - 0.27 = 0.62 (iv) PLA'n B) = P(B)- PLANB) [ which is clear from Venn = 0.49 -0.38 1 diagram) 0.11

(v) P(AUB) = PIA + PUB) - P(AnB) = [I-PIA)] + P(B) - PLANB) = I - PIA) + PIB) - PLANB) 2 - 0062 + 0.49 - 00) - 0.76 - cui P (BIA) - P(BNA) PIA) 0.1 T PLA) = \ - PLAT] 0:38 38