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.

(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.

= =

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.

The total no. of possible cases is 6^{3} = 216.

(i) Out of these 216, 1 - 2 - 3 occurs in 3! ways.

Probability = 3!/6^{3} =
1/36.

(ii) The no. of ways in which 11
is thrown is equal to the coeff. of x^{11} in (x + x^{2} +....+
x^{6})^{3} = 27.Probability = 27/216 = 1/8.

(iii) The favorable no. of cases =
sum of the coefficients of x^{3}, x^{4},... x^{10} in
the expansion of (x + x^{2} + .... x^{6})^{3}.

=
the coeff. of x^{10} in (x + x^{2} + .... +x^{6})^{3}
(1 + x + x^{2} + .....).

=
the coeff. of x^{10} in x^{3}(1 - x^{6})^{3} (1
- x)^{-4}.

=
the coeff. of x in (1 - 3x^{6}) (1 + 4x + x^{2}
+ x^{3} + .....)

= = 120 - 12 = 108

Probability of throwing less than 11 =

(Note : Here we made use of the fact that if S

=
a_{0} + a_{1}x + a_{2}x^{2} + .... then.

S(1
+ x + x^{2} + ....) = a_{0} + (a_{0} + a_{1})x
+ (a_{0} + a_{1} + a_{2})x^{2} + .......).

(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 – .

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.

Let E_{i} be the event that
the bag contains exactly i white balls

(0 £ i £
n)

&⇒ P(E_{i}) µ i

&⇒
P(E_{i}) = ki

&⇒ P(E_{0}) = 0.

Since {E_{i}} is the set of
mutually exclusive events.

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

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.

Let E_{i} (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(E_{i})= (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. E_{8} must occur P (E_{8}
/ A) =

1 ^{8}C_{6}
´ ^{12}C_{4
}=

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.

Let B_{i} 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(B_{i}) =

&⇒
P(B_{1}) = 1/42

P(B_{2}) = 5/21

P(B_{3}) = 10/21

P(B_{4}) = 5/21

P(B_{5}) = 1/42

Let ‘A’ be the probability that from the second bag a white ball is drawn

&⇒
P(A/B_{i}) =

&⇒ P(A) = SP(B_{i})×P(A/B_{i}) = 3/5

Now P(B_{2}/A) = = .