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PROBABILITY      579
Solution Let A, E1, E2, E3 and E4 be the events as defined below :
A : a black ball is selected
E1 : box I is selected
E2 : box II is selected
E3 : box III is selected
E4 : box IV is selected
Since the boxes are chosen at random,
Therefore
P(E1) = P(E2) = P(E3) = P(E4) = 1
4
Also
P(A|E1) = 3
18 , P(A|E2) = 2
8 , P(A|E3) = 1
7  and  P(A|E4) = 4
13
P(box III is selected, given that the drawn ball is black) = P(E3|A) By Bayes'
theorem,
P(E3|A) = 
3
3
1
1
2
2
3
3
4
4
P(E ) P(A|E )
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E )
P(E )P(A|E )
 
 
 
= 
1
1
4
7
0 165
1
3
1
1
1
1
1
4
4
18
4
4
4
7
4
13
×
=
×
+
×
+
×
+
×
Example 34 Find the mean of the Binomial distribution 
B 4, 31 Solution Let X be the random variable whose probability distribution is 
B 4,31