# Free Objective Test 01 Practice Test - 11th and 12th

The probability that A speaks truth is 45, while this probability for B is 34. The probability that they contradict each other when asked to speak on a fact is

A.

323

B.

120

C.

15

D.

1440

#### SOLUTION

Solution : D

In short the event described here is P(A)'P(B) +P(A)P(B)'

where P(A) and P(B) are the probabilities of A and B speaking the truth respectively.

P(E)= P(A)'P(B) +P(A)P(B)' = 15×34+45×14

P(E)=320+420=720=1440

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

A. 29
B. 218
C. 38
D. 120

#### SOLUTION

Solution : B

For a particular house being selected Probability =13

Prob(all the persons apply for the same house) =(13×13×13)×3=19 or 218

Ashu studies at Byju's classes and her probability of selection in IIT-JEE is 45. Ridhima took coaching at FIIT-JEE and the probability of her selection is 23. What is the probability that only 1 of them cracks the Exam?

A.

25

B.

310

C.

415

D.

None of the above.

#### SOLUTION

Solution : A

P(A)= 45

P(R)=23

P(E)= P(A'R)+P(AR')

P(E)=15×23+45×13

P(E)=215+415

P(E)= 615

P(E)= 25

Probability of a fraudster being caught is 12 and committing a fraud is 35. What is his chance of not going to jail?

A. 70%
B. 50%
C. 40%
D. 35%

#### SOLUTION

Solution : A

P(not going to jail)= P( F')+ P(F intersection C')

where P(F) is the probability of commiting a fraud and P(C) is the Probability of being caught

P(F')= 25

P(C')= 12

P(F intersection C')= P(F) × P(C')= 35×12=310

P(not going to jail)= 25+310=710=70%

There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, is a random order till both the faulty machines are identified. Then the probability that only two tests are needed

A. 13
B. 16
C. 12
D. 14

#### SOLUTION

Solution : B

This is a problem of without replacement.
P=one def. from 2 def.any one from 4×1 def. from remaining 1 def.any one from remaining 3
Hence required probability = 24×13=16
Aliter : Number of ways in which two faulty machines may be detected (depending upon the test done to identify the faulty machines) = 4C2=6
Number of favourable cases = 1
[When faulty machines are identified in the first and the second test].
Hence required probability = 16

Two persons A and B take turns in throwing a pair of dice. The first person to through 9 from both dice will be awarded the prize. If A throws first then the probability that B wins the game is

A. 917
B. 817
C. 89
D. 19

#### SOLUTION

Solution : B

The probability of throwing 9 with two dice = 436=19
The probability of not throwing 9 with two dice = 89
If A is to win he should throw 9 in 1st or 3rd or 5th attempt
If B is to win, he should throw, 9 in 2nd, 4th attempt
B’s chances = (89).19+(89)3.19+.....=89×191(89)2=817

In four schools B1,B2,B3,B4 the percentage of girls students is 12, 20, 13, 17 respectively. From a school selected at random, one student is picked up at random and it is found that the student is a girl. The probability that the school selected is B2, is

A. 631
B. 1031
C. 1362
D. 1762

#### SOLUTION

Solution : B

Favorable number of cases = 20C1=20
Sample space = 62C1=62
Required probability = 2062=1031

Six boys and six girls sit in a row. What is the probability that the boys and girls sit alternatively

A. 1462
B. 1924
C. 12
D. None of these

#### SOLUTION

Solution : A

Let n = total number of ways = 12!
and m = favourable numbers of ways = 2×6!.6!
Since the boys and girls can sit alternately in 6! . 6! ways if we begin with a boy and similarly
they can sit alternately in 6! . 6! Ways if we begin with a girl
Hence required probability  = mn=2×6!.6!12!=1462

Cards are drawn one by one at random from a well shuffled full pack of 52 cards until two aces are obtained for the first time. If N is the number of cards required to be drawn, then PrN=n where  2n50, is

A. (n1)(52n)(51n)50×49×17×3
B. 2(n1)(52n)(51n)50×49×17×3
C. 3(n1)(52n)(51n)50×49×17×3
D. 4(n1)(52n)(51n)50×49×17×3

#### SOLUTION

Solution : A

Here the least number of draws to obtain 2 aces are 2 and the maximum number is 50 thus n can take value from 2 to 50.
Since we have to make n draws for getting two aces, in (n – 1) draws, we get any one of the 4 aces and in the nth draw we get one ace. Hence the required probability
=4C1×48Cn252Cn1×352(n1)
=4×(48)!(n2)!(48n+2)!×(n1)!(52n+1)!(52)!×352n+1
=(n1)(52n)(51n)50×49×17×3 (on simplification).

Let X be a set containing n elements. If two subsets A and B of X are picked at random, the probability that A and B the same number of elements, is

A. 2nCn22n
B. 12nCn
C. 1.3.5....(2n1)2n
D. 3n4n

#### SOLUTION

Solution : A

We know that the number of sub-sets of a set containing n elements is 2n.
Therefore the number of ways of choosing A and B is 2n.2n=22n
We also know that the number of sub-sets (of X) which contain exactly r elements is nCr.
Therefore the number of ways of choosing A and B, so that they have the same number elements is
(nC0)2+(nC1)2+(nC2)2+....+(nCn)2=2nCn
Thus the required probability = 2nCn22n.

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

A. 12
B. 715
C. 215
D. 13

#### SOLUTION

Solution : B

The number of ways to arrange 7 white an 3 black balls in a row
10!7!.3!=10.9.81.2.3=120
Numbers of blank places between 7 balls are 6. There is 1 place before first ball and 1 place after last ball. Hence total number of places are 8.
Hence 3 black balls are arranged on these 8 places so that no two black balls are together in number of ways.
=8C3=8×7×61×2×3=56
So required probability = 56120=715

One of the two events must occur. If the chance of one is 23 of the other, then odds in favour of the other are

A. 2:3
B. 1:3
C. 3:1
D. 3:2

#### SOLUTION

Solution : D

Let p be the probability of the other event, then the probability of the first event is 23 p. Since two events are totally exclusive, we have p+(23)p=1p=35
Hence odds in favour of the other are 3 : 5 – 3, i.e, 3 : 2.

Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from urn A and placed in urn B. Then one ball is drawn at random from urn B and placed in urn A. If one ball is now drawn at random from urn A, the probability that it is found to be red, is

A. 3255
B. 2155
C. 1955
D. None of these

#### SOLUTION

Solution : D

Let the events are
R1 = A red ball is drawn from urn A and placed in B
B1 = A black ball is drawn from urn A and placed in B
R2 = A red ball is drawn from urn A and placed in A
B2 = A black ball is drawn from urn A and placed in B
R = A red ball is drawn in the second attempt from A
Then the required probability
=P(R1R2R)+(R1B2R)+P(B1R2R)+P(B1B2R)
=P(R1)P(R2)P(R)+P(R1)P(B2)P(R)+P(B1)P(R2)P(R)+P(B1)P(B2)P(R)
=610×511×610×610×611×510×410×411×710+410×711×610
=3255

If A and B are two events such that P(A) = 12 and P(B) = 23, then

A. P(AB)23
B. 16P(AB)12
C. 16P(AB)12
D. All of the above

#### SOLUTION

Solution : D

We have P(AB)           max. {P(A),P(B)}=23
P(AB)               min.  {P(A),P(B)}=12
and P(AB)=P(A)+P(B)P(AB)P(A)+P(B)1=16
16P(AB)12
P(AB)=P(B)P(AB)
Hence 2312P(AB)2316
16P(AB)12

The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays is ____.

A. 27
B. 47
C. 37
D. 17

#### SOLUTION

Solution : C

A leap year consists of 366 days comprising of 52 weeks and 2 days. There are 7 possibilities for these 2 extra days.
(i) Sunday, Monday
(ii) Monday, Tuesday
(iii) Tuesday, Wednesday
(iv) Wednesday, Thursday
(v) Thursday, Friday
(vi) Friday, Saturday
(vii) Saturday, Sunday
Let us consider two events :
A : the leap year contains 53 Sundays
B : the leap year contains 53 Mondays.
Then we have P(A)=27,P(B)=27,P(AB)=17
Required probability = P(AB)
=P(A)+P(B)P(AB)=27+2717=37