Each packet of certain items contains a coupon, which is equally likely to bear the letters A, N, S, H or U. If ‘m’ packets are purchased, find the probability that the coupons can not be used to spell ANSHU.

Let E_{1} is
the event that A is not present,

E_{2} is the event that N is not
present,

E_{3} is the event that S is not
present,

E_{4} is the event that H is not
present and

E_{5} is the event that U is not
present

Then required probability

=

=

= .

Three dice are rolled together till a sum of either 4 or 5 or 6 is obtained. Find the probability that 4 comes before 5 or 6.

We need (x + x^{2}
+ x^{3} + x^{4} + x^{5} + x^{6})^{3} =
x^{3}(1 - x^{6})^{3} (1 - x)^{-3}

= x^{3}(1 - 3x^{6} +
3x^{12} - x^{18}) (1 + 3x + 6x^{2} + 10x^{3}+...)
.

coefficient of x^{4} = 3, coefficient
of x^{5} = 6, coefficient of x^{6} = 10.

A bag contains n white and n black balls, all of equal size. Balls are drawn at random. Find the probability that there are both white and black balls in the draw and that the number of white balls is greater than that of black balls by one. .

Let S be the sample space consisting
of elements representing balls that can be drawn from the bag containing 2n
balls (n white + n black). Let E_{ij} be the event representing drawing
of i white and j black balls. Let E be the event of drawing balls such that
number of white balls is greater than that of black by one.

Then E = E_{21} ÈE_{32}
È
......... È E_{n}, _{n - 1}.

m (S) = ^{2n}C_{1} + ^{2n}C_{2}
+ ... + ^{2n}C_{2n} = 2^{2n} - 1.

m(E) = m(E_{21}) + m(E_{32})
+ ... + m(E_{n}, _{n - 1}).

= ^{n}C_{2} . ^{n}C_{1}
+ ^{n}C_{3} . ^{n}C_{2} + …. + ^{n}C_{n}
. ^{n}C_{n - 1} = ^{2n}C_{n - 1} – ^{n}C_{1}
. ^{n}C_{0} .

= ^{2n}C_{n - 1} – n.

Hence

Note:

m(S) stands for the total number of ways.

m (E) stands for favourable number of ways. .

In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choice (of only one is correct). The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it, is 1/8. Find the probability that he know the answer given that he answered it correctly. .

Let A_{1}, A_{2}, A_{3}
be event that examinee guesses, copies or knows the answer.

P(A_{1}) = 1/3, P(A_{2}) = 1/6, P(A_{3}) = ½

(as A_{1}, A_{2} and A_{3} are exhaustive)

Let A be one event that examinee answers one question correctly,

P(A/A_{1}) = 1/4, P(A/A_{2}) = 1/8, P(A/A_{3}) = 1

Now P(A) = P(A_{1}) . P(A/A_{1}) + P(A_{2}) . P(A/A_{2})
+ P(A_{3}) P(A/A_{3})

= =

Now P(A_{3}/A) = = .

There are two bags each containing 10 books all having different titles but of the same size. A student draws out any number of books from first bag as well as from the second bag. Find the probability that the difference between the number of books drawn from the two bags does not exceed two.

Let ‘S’ be the sample space, A_{0}
be the event that books drawn from two bags are equal in number, A_{1}
be the event that number of books drawn from one bag exceed those drawn from
another bag by one, and A_{2} be the event that number of books drawn
from one bag exceed those drawn from other bag by two.

Total ways = (^{10}C_{1}
+ ^{10}C_{2} + L
+ ^{10}C_{10})^{2} = (2^{10} - 1)^{2}

Favourable ways for A_{0},

= (^{10}C_{1})^{2}
+ (^{10}C_{2})^{2} + L + (^{10}C_{10})^{2}
= ^{20}C_{10}^{ }-
1

Favourable ways for A_{1},

= 2(^{10}C_{1}×^{10}C_{2} + ^{10}C_{2}×^{10}C_{3}^{ }+
L + ^{10}C_{9}×^{10}C_{10})

= 2(^{20}C_{9}^{ }- ^{10}C_{0}×^{10}C_{1}) = 2 (^{20}C_{9}
- 10)

Favourable ways for A_{2}

= 2(^{10}C_{1}.^{10}C_{3}
+ ^{10}C_{2}.^{10}C_{4} + L + ^{10}C_{8}.^{10}C_{10}).

= 2(^{20}C_{8} - ^{10}C_{0}.^{10}C_{2}).

= 2(^{20}C_{8} - 45)

&⇒
Required probability = .