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Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 4 of 78 pages
Question 31. Seven coupons are selected at random one at a time with replacement from 15 coupons numbered 1 to 15. The probability that the largest number appearing on a selected coupon is 9, is
  1.    (916)6
  2.    (815)7
  3.    (35)7
  4.    (35)7 - (815)7
 Discuss Question
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 9787157
=(35)7(815)7
Question 32. Three six faced fair dice are rolled together.  The probability that the sum of the numbers appearing on the dice is 8, is
  1.    772
  2.    754
  3.    427
  4.    764
 Discuss Question
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.
Question 33. CSK lost the toss 13 out of 14 times. What is the probability of this event?
  1.    3214
  2.    5214
  3.    7213
  4.    None of the above
 Discuss Question
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
Question 34. A set of integers is given as (3,6,8,14,17). What is the probability that a triangle can be constructed.?
  1.    35
  2.    310
  3.    110
  4.    510
 Discuss Question
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
Question 35. Letters of the word "EDUCATION" is arranged. What is the probability that vowels and consonants are in alphabetical order?
  1.    1(5!×4!)
  2.    19!
  3.    99!
  4.    None of these.
 Discuss Question
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!)
Question 36. A bag contains 'x' red balls '2x' white balls and '3x' black balls. 3 balls are drawn at random. The probability that all the balls drawn are of different colors is 0.3 .How many white balls are present in the bag?
  1.    2
  2.    4
  3.    5
  4.    7
 Discuss Question
Answer: Option A. -> 2
:
A
P(All different)= (xC1)×(2xC1)×(3xC1)/(6xC3)
P(E)= (6x3)×66x×[(6x1)×(6x2)]=310
or, 3x2[(6x1)×(6x2)]=310
10x2=18x29x+1
or, (x-1)(8x-1)=0
x can not be a fraction hence x=1
Total number of white balls= 2x=2
Question 37. There are 4 torn books in a lot consisting of 10 books. If 5 books are seleceted at random what is the probability of presence of 2 torn books among the selected?
  1.    13
  2.    421
  3.    1021
  4.    27
 Discuss Question
Answer: Option C. -> 1021
:
C
Total Sample space: 10C5=252
Favorable cases: 4C2×6C3=120
Hence P(E)= 120252=1021
Question 38. 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
  1.    2:3
  2.    1:3
  3.    3:1
  4.    3:2
 Discuss Question
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=1p=35
Hence odds in favour of the other are 3 : 5 – 3, i.e, 3 : 2.
Question 39. The probability of happening an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of happening neither A nor B is
  1.    0.6
  2.    0.2
  3.    0.21
  4.    None of these 
 Discuss Question
Answer: Option B. -> 0.2
:
B
P(¯¯¯¯A¯¯¯¯B)=P(¯¯¯¯A¯¯¯¯B)1P(AB)
Since A and B are mutually exclusive,
P(AB)=P(A)+P(B)
Hence required probability =1(0.5+0.3)=0.2
Question 40. 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
  1.    2:3
  2.    1:3
  3.    3:1
  4.    3:2
 Discuss Question
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=1p=35
Hence odds in favour of the other are 3 : 5 – 3, i.e, 3 : 2.

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