Question 970604
There were three candidates for the position of the chairman of a college Mr. X, Mr. Y and Mr. Z whose chances of getting appointments are in 
the ratio 4:2:3 respectively. The probability that Mr. X if selected would introduce computer education in the college is 0.3. The probabilities of Mr. Y and Mr. Z doing the same are 0.5 and 0.8 respectively. What is the
probability that there was computer education in the college?	

Solution:
First we compute the probability that there was NO computer education in the college.
P(C') = {{{(4/(4+2+3))(1-0.3) + (2/(4+2+3))(1-0.5) + (3/(4+2+3))(1-0.8)}}} = {{{22/45}}}
Therefore, the probability that there was computer education in the college is
P(C) = 1 - P(C') = 1 - 22/45 = {{{highlight(23/45)}}}