The CEO of a corporation must select 3 individuals to head the firm’s annual drive for community charities....?
There are three divisions (A, B, and C). Division A has 5 candidates, Division B has 6, and Division C has 4. If the CEO puts all the candidates' names into a had and draws 3 names out totally at random, what is the probability that the third name drawn is not from division A?
More Questions From This User See All
Answers & Comments
Verified answer
since the divisions are equally likely to turn up in any position,
P(last not from A) = P(1st not from A) = 10/15 = 2/3 <------
ps:
----
if you don't believe it is so simple, here's a full check
XXX: 10*9*8 = 720
AXX: 5*10*9 = 450
XAX: 10*5*9 = 450
AAX: 5*4*10 = 200 ....... 1820 favorable
AAA: 5*4*3 = 60
XAA: 10*5*4 = 200
AXA: 5*10*4 = 200
XXA: 10*9*5 = 450 ......910 unfavorable 2730 total
Pr = 1820/2730 = 2/3 <----