Quantitative Aptitude
PROBABILITY MCQs
Probability, Probability I
Total Questions : 775
| Page 4 of 78 pages
Answer: Option D. -> (35)7 - (815)7
:
D
Each coupon can be selected in 15 ways. The total number of ways of choosing 7 copouns is 157. If largest number is 9, then the selected numbers have to be from 1 to 9 excluding those consisting of only 1 to 8.
Probability desired is 97−87157
=(35)7−(815)7
:
D
Each coupon can be selected in 15 ways. The total number of ways of choosing 7 copouns is 157. If largest number is 9, then the selected numbers have to be from 1 to 9 excluding those consisting of only 1 to 8.
Probability desired is 97−87157
=(35)7−(815)7
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 C. -> 7213
:
C
13 losses from 14 can be selected in 14C13 ways i.e.14
At any of the tosses only 2 events can happen win or loss hence total sample space: 214
P(E)= 14214
:
C
13 losses from 14 can be selected in 14C13 ways i.e.14
At any of the tosses only 2 events can happen win or loss hence total sample space: 214
P(E)= 14214
Answer: Option B. -> 310
:
B
Any 3 points can be selected in 5C3 i.e 10 ways.
For forming a triangle sum of two sides should be greater than the 3rd side.
Hence following set of integers can be selected
(3,6,8)
(6,8,17)
(8,14,17)
Hence P(Triangle is formed)= 310
:
B
Any 3 points can be selected in 5C3 i.e 10 ways.
For forming a triangle sum of two sides should be greater than the 3rd side.
Hence following set of integers can be selected
(3,6,8)
(6,8,17)
(8,14,17)
Hence P(Triangle is formed)= 310
Answer: Option A. -> 1(5!×4!)
:
A
There are 5 vowels and 4 consonants.
Total ways to arrange: 9!
Vowels can be arranged in 5! ways out of which in 15! they are in alphabetical order.
Similarly for consonants.
Total Favorable ways: 9!(5!×4!)
Total ways: 9!
P(E)= 1(5!×4!)
:
A
There are 5 vowels and 4 consonants.
Total ways to arrange: 9!
Vowels can be arranged in 5! ways out of which in 15! they are in alphabetical order.
Similarly for consonants.
Total Favorable ways: 9!(5!×4!)
Total ways: 9!
P(E)= 1(5!×4!)
Answer: Option A. -> 2
:
A
P(All different)= (xC1)×(2xC1)×(3xC1)/(6xC3)
P(E)= (6x3)×66x×[(6x−1)×(6x−2)]=310
or, 3x2[(6x−1)×(6x−2)]=310
10x2=18x2−9x+1
or, (x-1)(8x-1)=0
x can not be a fraction hence x=1
Total number of white balls= 2x=2
:
A
P(All different)= (xC1)×(2xC1)×(3xC1)/(6xC3)
P(E)= (6x3)×66x×[(6x−1)×(6x−2)]=310
or, 3x2[(6x−1)×(6x−2)]=310
10x2=18x2−9x+1
or, (x-1)(8x-1)=0
x can not be a fraction hence x=1
Total number of white balls= 2x=2
Answer: Option C. -> 1021
:
C
Total Sample space: 10C5=252
Favorable cases: 4C2×6C3=120
Hence P(E)= 120252=1021
:
C
Total Sample space: 10C5=252
Favorable cases: 4C2×6C3=120
Hence P(E)= 120252=1021
Answer: Option D. -> 3:2
:
D
Let p be the probability of the other event, then the probability of the first event is 23p. 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.
:
D
Let p be the probability of the other event, then the probability of the first event is 23p. 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.
Answer: Option B. -> 0.2
:
B
P(¯¯¯¯A∩¯¯¯¯B)=P(¯¯¯¯A∪¯¯¯¯B)−1−P(A∪B)
Since A and B are mutually exclusive,
P(A∪B)=P(A)+P(B)
Hence required probability =1–(0.5+0.3)=0.2
:
B
P(¯¯¯¯A∩¯¯¯¯B)=P(¯¯¯¯A∪¯¯¯¯B)−1−P(A∪B)
Since A and B are mutually exclusive,
P(A∪B)=P(A)+P(B)
Hence required probability =1–(0.5+0.3)=0.2
Answer: Option D. -> 3:2
:
D
Let p be the probability of the other event, then the probability of the first event is 23p. 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.
:
D
Let p be the probability of the other event, then the probability of the first event is 23p. 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.