Quantitative Aptitude
PROBABILITY MCQs
Probability, Probability I
Total Questions : 775
| Page 3 of 78 pages
Answer: Option D. -> 0.001
:
D
The probability of anevent always lies between 0 and 1 (0 and 1 inclusive).
Hence, among the options, only 0.001 can be the probability of an event.
:
D
The probability of anevent always lies between 0 and 1 (0 and 1 inclusive).
Hence, among the options, only 0.001 can be the probability of an event.
Answer: Option A. -> No. of favourable outcomes / Total number of outcomes
:
A
.
:
A
.
:
Out of the givne 5 bags, bag 2 and bag 4 have more than 18 chocolates.
Probability(E) =number of favorable eventstotal number of events
∴ Probablity of a bag containing more than 18 chocolates =number of bags containing more than 18 chocolatestotal number of bags=25=0.4
Answer: Option B. -> 0 to 1 (inclusive)
:
B
.
:
B
.
Answer: Option B. -> 311
:
B
Total number of alphabets in the wordMISSIMMIPPI = 11
Number of times M comes = 3
Probability of picking M =Number of times M comesTotal number of alphabets in the word MISSIMMIPPI
So, probability of its occurrence will be 311.
:
B
Total number of alphabets in the wordMISSIMMIPPI = 11
Number of times M comes = 3
Probability of picking M =Number of times M comesTotal number of alphabets in the word MISSIMMIPPI
So, probability of its occurrence will be 311.
Answer: Option B. -> The probability that A wins and loses equal number of matches is 1781
:
B
Probability of equal number of W and L is the probability of following
(0)W,(0)L +(1)W,(1)L+(2)W, (2)L
=(13)5+5C1.4C1(13)5+5C2.3C2(13)5=1781
:
B
Probability of equal number of W and L is the probability of following
(0)W,(0)L +(1)W,(1)L+(2)W, (2)L
=(13)5+5C1.4C1(13)5+5C2.3C2(13)5=1781
Answer: Option A. -> 772
:
A
Getting sum 8 when three dice are rolled = E
The possibilities are (1,1,6) , (2,2,4), (3,3,2), (1,2,5), (1,3,4)
The number of favorable cases for E = 3 + 3 + 3 + 6 + 6 = 21
The total number of cases = 63= 216
Required probability P(E) =21216=772.
:
A
Getting sum 8 when three dice are rolled = E
The possibilities are (1,1,6) , (2,2,4), (3,3,2), (1,2,5), (1,3,4)
The number of favorable cases for E = 3 + 3 + 3 + 6 + 6 = 21
The total number of cases = 63= 216
Required probability P(E) =21216=772.
Answer: Option A. -> A little less than 12
:
A
Out of 9 distinct black and 9 distinct white balls, probability of drawing a white ball =12
and drawing black ball is also 12. for at least 4 of each colour in 9 draws with replacement, there are two cases,
Case(i)
P(getting 5 white, 4 black) =9C5(12)9
Cases (ii)
P(getting 4 white, 5 black)=9C4(12)9
These are exclusive so
P(atleast 4 of each colour) =9C5(12)9+9C4(12)9=63128 which is little less than 12
:
A
Out of 9 distinct black and 9 distinct white balls, probability of drawing a white ball =12
and drawing black ball is also 12. for at least 4 of each colour in 9 draws with replacement, there are two cases,
Case(i)
P(getting 5 white, 4 black) =9C5(12)9
Cases (ii)
P(getting 4 white, 5 black)=9C4(12)9
These are exclusive so
P(atleast 4 of each colour) =9C5(12)9+9C4(12)9=63128 which is little less than 12
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 nthdraw 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).
:
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 nthdraw 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).
Answer: Option A. -> 181
:
A
∑8x=0P(X)=1⇒P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)=1=λ+3λ+5λ+7λ+9λ+11λ+13λ+15λ+17λ=1⇒92(λ+17λ)=1⇒81λ=1∴λ=181
:
A
∑8x=0P(X)=1⇒P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)=1=λ+3λ+5λ+7λ+9λ+11λ+13λ+15λ+17λ=1⇒92(λ+17λ)=1⇒81λ=1∴λ=181