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

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

Three persons A_{1}, A_{2} and A_{3}
are to speak at a function along with 5 other persons. If the person speak in
random order, the probability that A_{1} speaks before A_{2}
and A_{2} speaks before A_{3} is’.

(A) 1/6

(B) 3/5

(C) 3/8

(D) none of these

Total number of ways in which 8 persons can speak is 8!

Now 3 positions out
of 8 position can be chosen in ^{8}C_{3} i.e. 56 ways and at
these positions we can put A_{1}, A_{2} and A_{3} in
the required order. Further the remaining persons can speak in 5! ways

&⇒ Total number of
favourable ways

= 56 (5! )

&⇒ Required
probability = = 1/6.

A fair die is tossed eight times. Probability that on the eighth throw a third six is observed is, .

(A) ^{8}C_{3}

(B)

(C)

(D) none of these

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 = ^{7}C_{2} . (1/6)^{2}. (5/6)^{5}.
(1/6)

= .

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) 15/2^{8} .

(B) 2/15

(C) 15/2^{13}

(D) none of these

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} C_{7}(1/2)^{7} (1/2)^{n-7} = ^{n}C_{9}(1/2)^{n-9}
´ (1/2)^{9}

&⇒ ^{n}C_{7}
= ^{n}C_{9}

&⇒
n = 16

Now P (X = 2) = ^{16}C_{2}(1/2)^{2}
(1/2)^{14}

= = .

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

Total number of
ways in which 4 papers can be distributed among 7 teachers = 7^{4}
.

Now exactly 2
teachers out of 7 can be chosen in ^{7}C_{2} ways. And total
number of ways in which 4 papers can be given to these 2 teachers (
each one getting atleast one) = (2^{4} –2 ) =14.

&⇒ Total number of ways in which
exactly 2 teachers check all four papers = ^{7}C_{2}. 14=
21 .14.

&⇒ Required probability =

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

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

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

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 = =

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 –

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– .

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

Probability that out of remaining balls the one that is red is

=

If
‘head’ means one and ‘tail’ means two , then coefficient of quadratic equation
ax^{2} + 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

b^{2}
– 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