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
Let E1 is
the event that A is not present,
E2 is the event that N is not
E3 is the event that S is not
E4 is the event that H is not
E5 is the event that U is not
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
We need (x + x2
+ x3 + x4 + x5 + x6)3 =
x3(1 - x6)3 (1 - x)-3
= x3(1 - 3x6 +
3x12 - x18) (1 + 3x + 6x2 + 10x3+...)
coefficient of x4 = 3, coefficient
of x5 = 6, coefficient of x6 = 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 Eij 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 = E21 ÈE32
......... È En, n - 1.
m (S) = 2nC1 + 2nC2
+ ... + 2nC2n = 22n - 1.
m(E) = m(E21) + m(E32)
+ ... + m(En, n - 1).
= nC2 . nC1
+ nC3 . nC2 + …. + nCn
. nCn - 1 = 2nCn - 1 – nC1
. nC0 .
= 2nCn - 1 – n.
m(S) stands for the total number of
m (E) stands for favourable number of
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 A1, A2, A3
be event that examinee guesses, copies or knows the answer.
P(A1) = 1/3, P(A2) = 1/6, P(A3) = ½
(as A1, A2 and A3 are exhaustive)
Let A be one event that examinee answers one question correctly,
P(A/A1) = 1/4, P(A/A2) = 1/8, P(A/A3) = 1
Now P(A) = P(A1) . P(A/A1) + P(A2) . P(A/A2)
+ P(A3) P(A/A3)
Now P(A3/A) = = .
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, A0
be the event that books drawn from two bags are equal in number, A1
be the event that number of books drawn from one bag exceed those drawn from
another bag by one, and A2 be the event that number of books drawn
from one bag exceed those drawn from other bag by two.
Total ways = (10C1
+ 10C2 + L
+ 10C10)2 = (210 - 1)2
Favourable ways for A0,
+ (10C2)2 + L + (10C10)2
= 20C10 -
Favourable ways for A1,
= 2(10C1×10C2 + 10C2×10C3 +
L + 10C9×10C10)
= 2(20C9 - 10C0×10C1) = 2 (20C9
Favourable ways for A2
+ 10C2.10C4 + L + 10C8.10C10).
= 2(20C8 - 10C0.10C2).
= 2(20C8 - 45)
Required probability = .