Probability I(11th And 12th > Mathematics ) Questions and Answers
Explanation:-
Answer: Option D. -> 1440:
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
Explanation:-
Answer: Option A. -> 25:
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
Explanation:-
Answer: Option A. -> 70%:
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%
Explanation:-
Answer: Option B. -> 218:
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
Explanation:-
Answer: Option B. -> 817:
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
Explanation:-
Answer: Option B. -> 16:
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
Explanation:-
Answer: Option B. -> 1031:
B
Favorable number of cases = 20C1=20
Sample space = 62C1=62
∴ Required probability = 2062=1031
Explanation:-
Answer: Option A. -> (n−1)(52−n)(51−n)50×49×17×3:
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).
Explanation:-
Answer: Option A. -> 1462:
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
Explanation:-
Answer: Option A. -> 2nCn22n:
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.