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