1. In how many
different ways can the letters of the word ‘DETAILS’ be arranged in such a way
that the vowels occupy only the odd
positions?
a) 32 b) 48 c) 36
d) 60
Sol : There are 6 letters in the given word, out
of wich there are 3 vowels and 3
consonants.
3
vowels can be placed at any of the 3 places
marked 1, 3, 5.
No of ways of arranging the vowels = 3P3 = 3! =6
Also the 3 consonants can be arranged at the remaining 3 positions.
No of ways these arrangements = 3P3 = 3! = 6
Total no of ways = 6*6 =36
2. From a group of 7
men and 6 woman five persons are to be
selected to form a committee so that at least 3 men are there on te committee .
In how many ways can it be done?
a)
564 b) 645 c)
735 d) 756
Sol : We may have
(3 men and 2 women ) or (4 men and 1 woman ) or(5 men only)
Required no of ways = 7C3 * 6C2 +
7C4 * 6C1 +7C5
= 756
3. A box contains 2 white balls , 3 black balls and 4
red balls . In how many ways can 3 balls be
drawn from
the box, if at least one black ball is to be included in the draw?
a) 32 b) 48 c) 64 d) 96
Sol: we may
have (1 black & 2 non-black ) or (2 black and 1 non black ) or 3 black
Required no of ways = 3C1 * 6C2 + 3C2 *
6C1 + 3C3
= 45+18+1=64
4. Two dice are thrown together . What is the probability that the sum of the nos on
the two
faces is
divisible by 4 or 6?
a) 7/18 b) 5/ 18
c) 5/36 d) ½
Sol : n(S) = 6 * 6 =36
Let E be the event tat the sum of the nos on the two faces is divisible
by 4 or 6 Then
E
= {(1,3) ,(1, 5) , (2, 2), (2, 4) ,(2, 6), (3,1), (3,3) , (3,5), (4,2), (4,4),
(5,1), (5,3),(6,2),
(6,6)}
n(E)
= 14
P(E) = n(E)/n(S) = 14/ 36 = 7/18
5. A box contains 5
green, 4 yellow and 3 white marbles. Three marbles are drawn at
random. What
is the probability that they are not of the same colour?
a) 3/44 b)3/55 c) 52/55 d)41/44
Sol: Let S be the sample space. Then n(S) = no of
ways of drawing 3 marbles out of 12.
=
12C3 = 220
Let E be the event of drawing 3 balls of the same colour.
Then E = events of drawing (3 balls out of 5) or (3 balls out of 4 ) or
(3 balls out of 3)
ð
n(E) = 5C3 +4C3 +3C3 = 5C2 + 4C1 + 1 = 15
ð
P(E) = n(E)/n(S) = 15/220 = 3/44
Required probability = 1-(3/44) =41/44
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