Of the 8 members of flight crew, 7 are trained to serve economic class customer, 5 are trained to serve business class customer, and 4 can serve both. If one member is chosen at random, what is the probability that the member chosen will be someone who can serve either type of customer?	Options: A) a) 1/9 B) b) 0 C) c) 1/6 D) d) 1/4 E) e) 1/3
You know how to solve for Total number of people who can serve at least one type of customer:   Total = n(A) + n(B) - n(A and B)   Total people who can serve at least one type of customer = 7 + 5 - 4= 
<gadget id="calculator">7 + 5 - 4</gadget>
<output>8</output>
8   So All 8 people know how to serve at least one type of customer. Hence, if you pick anyone, the probability is 0 that he/she wouldn't know how to serve any type of customer   Answer (b) 0)
<result>B</result>
Of the 8 members of flight crew, 7 are trained to serve economic class customer, 5 are trained to serve business class customer, and 4 can serve both. If one member is chosen at random, what is the probability that the member chosen will be someone who can serve either type of customer?