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

### Question 1

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

323

120

15

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

### Question 2

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

#### 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

### Question 3

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?

25

310

415

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

### Question 4

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

#### 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%

### Question 5

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

#### 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

### Question 6

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

#### 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

### Question 7

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

#### SOLUTION

Solution :B

Favorable number of cases = 20C1=20

Sample space = 62C1=62

∴ Required probability = 2062=1031

### Question 8

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

#### 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

### Question 9

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 2≤n≤50, is

#### 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

=4C1×48Cn−252Cn−1×352−(n−1)

=4×(48)!(n−2)!(48−n+2)!×(n−1)!(52−n+1)!(52)!×352−n+1

=(n−1)(52−n)(51−n)50×49×17×3 (on simplification).

### Question 10

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

#### 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.

### Question 11

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

#### 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

### Question 12

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

#### 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=1⇒p=35

Hence odds in favour of the other are 3 : 5 – 3, i.e, 3 : 2.

### Question 13

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

#### 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

### Question 14

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

#### SOLUTION

Solution :D

We have P(A∪B)≥ max. {P(A),P(B)}=23

P(A∩B)≤ min. {P(A),P(B)}=12

and P(A∩B)=P(A)+P(B)−P(A∪B)≥P(A)+P(B)−1=16

⇒16≤P(A∩B)≤12

P(A′∩B)=P(B)−P(A∩B)

Hence 23−12≤P(A′∩B)≤23−16

⇒16≤P(A′∩B)≤12

### Question 15

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

#### 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(A∩B)=17

∴ Required probability = P(A∪B)

=P(A)+P(B)−P(A∪B)=27+27−17=37