Question

I want to see the thought process behind this so that I could approach similar problems on my own, so please attempt to show every step and formula used, thank you.

6. At a company, 70% of the employees have License 1, 60% have License 2, 20% have neither. Find the probability that a selected employee has (i) License 1 but not License 2. (ii) both licenses. (ii) License 2 given that the employee has License 1.

Answer 1

TOPIC:Events and probability.

6) Define the following Ex : Having 22 : Having events as- License ; License 2. . Given that, P (of having License )) = 0.70 e (21) = 0.70 & (of having License 2 = 0.60 P (22) = 0.60) Also, P. (of having neither licensed 20.20 - P ( n ) = 0.20. =) P (2, 042) = 0.20. [ De- mors gan's] law. eston 1-P (2, v. 22) = 0.20. I Law of complemen- tation = P (2, 0 42) =0.80. o n -0.80 [ Additive law of n - P (21) +P (22) -P (2, 22) = 0.80. I proobability - 0.70 +0.60 - P(in ) = 0.80. = [ (in a) = 0.50

we need =) 8 (dicense I but not License 2). = (404). 7 P (21) - P (41 en 42). P (4) = 0.70, P (2,04₂) = 0,50. = 0.70 - 0.50. 10. 20) (An). (Link) [= 2,- (2.142] . We need, & (of having both licenses. =P (of having license 1 and license 2). =P (2, n. 42). = 10.50 (Anse) ( We needs P (License 2) The employee has License Y. =P (22/4), P (2204 I Conditional law) P (21) of probability, - Car [commutative lan - 0.50 = 10.7143 cans Pan 22) I commutative law 0.50 0.70