Question.1
A fair die is thrown
until a score of less than 5 points is obtained. The probability of obtaining
not less than 2 points on the last thrown is.
(A) 3/4
(B) 5/6
(C) 4/5
(D) 1/3
Solution
Score less than 5 means the occurrence of 1, 2, 3, or
4. Now on the last throw we should not obtain a score less then 2 i.e. one.
Clearly the favourable outcomes are 2, 3 or 4. .
Thus the required
probability = 3/4
Question.2
Three persons A1, A2 and A3
are to speak at a function along with 5 other persons. If the person speak in
random order, the probability that A1 speaks before A2
and A2 speaks before A3 is’.
(A) 1/6
(B) 3/5
(C) 3/8
(D) none of these
Solution
Total number of ways
in which 8 persons can speak is 8!
Now 3 positions out
of 8 position can be chosen in 8C3 i.e. 56 ways and at
these positions we can put A1, A2 and A3 in
the required order. Further the remaining persons can speak in 5! ways
&⇒ Total number of
favourable ways
= 56 (5! )
&⇒ Required
probability = 
= 1/6.
Question.3
A fair die is tossed
eight times. Probability that on the eighth throw a third six is observed is, .
(A) 8C3
(B) 
(C) 
(D) none of these
Solution
Third six occurs on
8th trial. It means that in first 7 trials we must exactly 2 sixes and 8th
trial must result in a six.
&⇒ Required
probability = 7C2 . (1/6)2. (5/6)5.
(1/6)
= 
.
Question.4
A fair coin is
tossed a fixed number of times . If the probability of getting 7 heads is
equal to getting 9 heads, then the probability of getting 2 heads is,
(A) 15/28 .
(B) 2/15
(C) 15/213
(D) none of these
Solution
Let coin was tossed 'n' times and X be the
random variable representing the number of head appearing in 'n' trials.
P(X = 7) = P (X = 9) .
&⇒
n C7(1/2)7 (1/2)n-7 = nC9(1/2)n-9
´ (1/2)9
&⇒ nC7
= nC9
&⇒
n = 16
Now P (X = 2) = 16C2(1/2)2
(1/2)14
= 
= 
.
Question.5
If the papers of 4 students can be checked by any one
of the 7 teachers, then the probability that all the 4papers are checked
by exactly 2 teachers is;
(A) 2/ 7
(B) 32/ 343
(B) 12/ 49
(D) None of these
Solution
Total number of
ways in which 4 papers can be distributed among 7 teachers = 74
.
Now exactly 2
teachers out of 7 can be chosen in 7C2 ways. And total
number of ways in which 4 papers can be given to these 2 teachers (
each one getting atleast one) = (24 –2 ) =14.
&⇒ Total number of ways in which
exactly 2 teachers check all four papers = 7C2. 14=
21 .14.
&⇒ Required probability = 
Question.6
Let 'E' and 'F' be two independent events. The probability
that both 'E’ and 'F’ happen is 1/12 and the probability that neither 'E' nor
'F' happens is ½, then ,.
(A) P(E) = 1/3, P(F) = 1/4
(B) P(E) = 1/2, P(F) = 1/6
(C) P(E) = 1/6, P(F)
= 1/2
(D) P(E) = 1/4, P(F)
= 1/3
Solution
P(EÇF) = P(E).P(F) = 1/12.
P(E¢ÇF¢) = P(E¢).P(F¢) .
&⇒ 1/2 = ( 1 – P(E) ) ( 1- P(F))
&⇒ P( E).P(F) + 1 – P(E) – P(F) = 1/2.
&⇒ P(E) + P(F) = 
&⇒ P(E) = 1/3, P(F) = 1/4
or P(E) = 1/4, P(F)
= 1/3
Question.7
There are n persons
(n ≥ 3), among whom are
A and B, who are made to stand in a row in random order. Probability that
there is exactly one person between A and B is .
(A) 
(B) 
(C) 2/n
(D) none of these
Solution
Person that must
stand between A and B can be chosen in (n-2) ways. Now number of ways in which
x person can be made stand so that there is exactly one person in between A and
B is equal to (n-2) . 2. (n-2)! .
Also total number of
ways in which persons can be made to stand =n!
&⇒ Required
probability = 
= 
Question.8
A number is chosen at random from the numbers 10 to 99.
By seeing the number a man will laugh if product of the digits is 12. If he
choose three numbers with replacement then the probability that he will laugh
at least once is .
(A) 1 – 
(B) 
(C) 1 – 
(D) 1 –
Solution
There can be four such numbers i.e. 43, 34, 62, 26.
Whose product of digit is 12.
&⇒ Probability that the man will laugh by seeing the
chosen numbers
= 
&⇒ Required probability = 1– 
.
Question.9
In a bag there are 15 red and 5 white balls. Two balls are
chosen at random and one is found to be red. The probability that the second
one is also red is .
(A) 12/19
(B) 13/19
(C) 14/19
(D) 15/19
Solution
Probability that out of remaining balls the one that is
red is
= 
Question.10
If
‘head’ means one and ‘tail’ means two , then coefficient of quadratic equation
ax2 + bx + c = 0 are chosen by tossing three fair coins. The
probability that roots of the equations are imaginary is .
(A)
5/8
(B)
3/8
(C)
7/8
(D)
1/8
Solution
b2
– 4ac < 0
For
b =1any a and c which can be chosen in 4 ways
For
b =2 either a = 1, c = 2
or
a = 2, c =1
or
a = 2, c = 2
&⇒
Required probability = 7/8
