Example.1
Suppose the manufacturing company in
Problem–13 decided it’s acceptance scheme. Under the new scheme an inspector
takes one item at random, inspects it, and then replaces it in the box; a
second inspector does likewise. Finally a third inspector goes through the same
procedure. The box is not shipped if any of the three find a defective. Answer
(a) and (b) in Problem 13 under this new scheme.
Solution
(a) A box containing three defectives will
be shipped, if all three inspectors don’t pick the defective item,
&⇒ Required probability = 
(b) A box containing
only one defective will be held if any one of the inspectors pick the defective
item.
&⇒ Required probability
= 1 - probability that no inspector picks
the defective item.
=
= 
Example.2
In a single throw with 3 dice, what is
the chance of throwing
(i) one-two-three
(ii) eleven
(iii) less than eleven
(iv) more than ten.
Solution
The total no. of possible cases is 63 = 216.
(i) Out of these 216, 1 - 2 - 3 occurs in 3! ways.
Probability = 3!/63 =
1/36.
(ii) The no. of ways in which 11
is thrown is equal to the coeff. of x11 in (x + x2 +....+
x6)3 = 27.Probability = 27/216 = 1/8.
(iii) The favorable no. of cases =
sum of the coefficients of x3, x4,... x10 in
the expansion of (x + x2 + .... x6)3.
=
the coeff. of x10 in (x + x2 + .... +x6)3
(1 + x + x2 + .....).
=
the coeff. of x10 in x3(1 - x6)3 (1
- x)-4.
=
the coeff. of x in (1 - 3x6) (1 + 4x +
x2
+
x3 + .....)
=
= 120 - 12 = 108
Probability
of throwing less than 11 = 
(Note
: Here we made use of the fact that if S
=
a0 + a1x + a2x2 + .... then.
S(1
+ x + x2 + ....) = a0 + (a0 + a1)x
+ (a0 + a1 + a2)x2 + .......).
(iv) The total no. of cases are
those which are either less than 11 or greater than equal to 11 (i.e. greater
than 10).
Thus required probability = 1 - prob.
of less than 11 = 1 –
.
Example.3
A bag contains n (white and black) balls.
It is given that the probability that the bag contains exactly r white balls is
directly proportional to r (0 £ r £ n). A ball is drawn at random and is
found to be white. Find the probability that there is only one white ball in
the bag.
Solution
Let Ei be the event that
the bag contains exactly i white balls
(0 £ i £
n)
&⇒ P(Ei) µ i
&⇒
P(Ei) = ki
&⇒ P(E0) = 0.
Since {Ei} is the set of
mutually exclusive events.

Let B be the event that the ball,
drawn out, is white

Problem.4
A bag contains a total of 20 books on
physics and Mathematics. Any possible combination of books is equally likely.
10 Books are chosen from the bag and it is found that it contains 6 books of
mathematics. Find out the probability that the remaining books in the bag
contains 2 books on mathematics.
Solution
Let Ei (i = 0,1,2,…, 20) be the
event that bag contains i books on mathematics. Since all these events are
equally likely and mutually exclusive and exhaustive so, P(Ei)=
(i = 0,1,2,.., 20) and let A be the
event that a draw of 10 books contains 6 books on mathematics.
P(A) = 
=
=

Now, we want the bag should
contain 2 more books on mathematics i.e. E8 must occur P (E8
/ A) = 
1 8C6
´ 12C4
=
Example.5
Consider
a bag containing 10 balls of which 4 are black. Now 5 balls are drawn from this
bag and put in another bag (without noting the colour of balls). Finally one
ball is drawn from the second bag and it is found to be white. Find the
probability that 2 white balls had been drawn in the first draw.
Solution
Let Bi represent the event that
in the first draw i (1 £ i £ 5) white balls and (5 - i) black balls had been drawn in the
first draw.
&⇒
P(Bi) = 
&⇒
P(B1) = 1/42
P(B2) = 5/21
P(B3) = 10/21
P(B4) = 5/21
P(B5) = 1/42
Let ‘A’ be the probability that from the
second bag a white ball is drawn
&⇒
P(A/Bi) = 
&⇒ P(A) = SP(Bi)×P(A/Bi) = 3/5
Now P(B2/A) =
=
.