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Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 3 of 78 pages
Question 21. Which of the following can be the probability of an event?
  1.    21/20
  2.    -0.5
  3.    3
  4.    0.001
 Discuss Question
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.
Question 22. Probability of occurrence of an event is defined as ______.
  1.    No. of favourable outcomes / Total number of outcomes
  2.    Favourable outcomes X Total number of outcomes.
  3.    Total number of outcomes / No. of favourable outcomes 
  4.    Happening + Not Happening 
 Discuss Question
Answer: Option A. -> No. of favourable outcomes / Total number of outcomes
:
A
.
Question 23. BagChocolates112219318422511
5 bags had some amount of chocolates in them, as shown in the table. The probability of occurrence of more than 18 chocolates in a bag if a bag is chosen at random will be ___
 Discuss Question

:
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
Question 24. The range of probability is
  1.    -1 to 1 (inclusive)
  2.    0 to 1 (inclusive)
  3.    1 to 2 (inclusive)
  4.    -1 to 0 (inclusive)
 Discuss Question
Answer: Option B. -> 0 to 1 (inclusive)
:
B
.
Question 25. If an alphabet is picked from the word MISSIMMIPPI what is the probability that an alphabet 'M' is picked?
  1.    111
  2.    311
  3.    811
  4.    1
 Discuss Question
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.
Question 26. Team A plays with 5 other teams exactly once. Assuming that for each match the probabilities of a win, draw and loss are equal, then
  1.    The probability that A wins and loses equal number of matches is 3481
  2.    The probability that A wins and loses equal number of matches is 1781
  3.    The probability that A wins more number of matches than it loses is 1781
  4.    The probability that A loses more number of matches than it wins is 1681
 Discuss Question
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
Question 27. 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 28. From a bag containing 9 distinct white and 9 distinct black,  9 balls are drawn at random one by one,  the drawn balls being replaced each time.  The probability that at least four balls of each colour is in the draw, is
  1.    A little less than 12
  2.    a little greater than 12
  3.    12
  4.    13
 Discuss Question
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
Question 29. 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  2n50, is
 
  1.    (n−1)(52−n)(51−n)50×49×17×3
  2.    2(n−1)(52−n)(51−n)50×49×17×3
  3.    3(n−1)(52−n)(51−n)50×49×17×3 
  4.    4(n−1)(52−n)(51−n)50×49×17×3
 Discuss Question
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
=4C1×48Cn252Cn1×352(n1)
=4×(48)!(n2)!(48n+2)!×(n1)!(52n+1)!(52)!×352n+1
=(n1)(52n)(51n)50×49×17×3(on simplification).
Question 30. A random variable X has the following probability distribution
XP(X=x)XP(X=x)0λ511λ13λ613λ25λ715λ37λ817λ49λ
then, λ is equal to 
 
  1.    181
  2.    281
  3.    581
  4.    781   
 Discuss Question
Answer: Option A. -> 181
:
A
8x=0P(X)=1P(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=192(λ+17λ)=181λ=1λ=181

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