Probability- Aptitude Questions and Answers - in Hindi

View in English View in Hindi

Find the probability that a leap year selected at random will contain 53 Sundays.


अज्ञात रूप से चुने गये लीप वर्ष मे 53 रविवार होने की संभावना ज्ञात करें।

A)
5/7

B)
3/4

C)
4/7

D)
2/7


Correct Answer :

2/7


Explanation :

Total number of days in a leap year = 366
It will contain 52 full weeks 2 extra days
thus sample space is n(S) : {sun-mon, mon-tue, tue-wed, wed-thu, thu-fri, fri-sat, sat-sun} = 7
out of these only two outcomes n(A): { sat-sun} and {sun-mon} is having sunday with them = 2

Therefore, probability that a leap year has 53 Sundays = n(A)/n(S) = 2/7

There are 5 red and 7 blue balls in a bucket. If one ball is drawn at random from the bucket then what is the probability that the ball drawn will be of red colour?


एक बाल्टी में 5 लाल और 7 नीली गेंदों है। यदि हम उसमें से एक गेंद बाहर निकालते है तो उसके लाल रंग की होने की क्या संभावना है?

A)
5/12

B)
7/12

C)
1/2

D)
1/12


Correct Answer :

5/12


Explanation :

total numbers of balls in bucket = 5 red+7 blue=12
sample space n(S)=number of ways of drawing one ball out of 12=12C1=>12
event of drawn one red ball out of 5 red balls n(E)=5C1=>5

hence probability p(E)=n(E)/n(S)=5/12

Two dices are thrown randomly. Find the probability of getting the sum of digits as 9.


दो पासों को अज्ञात रूप से फेका गया है। दोनो पासों पर आने वाली संख्याओं का योग 9 होने की संभावना ज्ञात करो।

A)
1/6

B)
1/9

C)
1/12

D)
None of these


Correct Answer :

1/9


Explanation :

All possible outcome when two dices are thrown randomly =sample space is n(S)=(6*6)=36 outcomes
event of getting the sum of digits as 9 are n(E) ={(3,6),(4,5),(5,4),(6,3)}=4 outcomes

Therefore Probability P(E)=n(E)/n(S)=4/36=1/9

A bag contains 2 red marbles, 3 green marbles and 4 blue marbles. A marble is picked at random from the bag. What is the probability that the marble is not a blue marble?


एक बैग में 2 लाल मार्बल, 3 हरे मार्बल और 4 नीले मार्बल है । बैग में से अज्ञात रुप से एक मार्बल निकाला जाता है। मार्बल का नीला मार्बल ना होने की क्या संभावना है?

A)
4/9

B)
5/9

C)
1/9

D)
2/3


Correct Answer :

5/9


Explanation :

total numbers of marbles in bag = 2 red+3 green+4 blue=9
sample space n(S)=number of ways of drawing one marble out of 9 marble from bag=9C1=>9
event of picking one marble which is not a blue
that is picking one marble from out of (2 red+3 green) 5 marbles, n(E)=5C1=>5

hence probability p(E)=n(E)/n(S)=5/9

A coin is tossed. What is the probability of getting a head or a tail?


एक सिक्‍का उछाला गया है । हेड या टेल आने की सम्‍भावना क्या होगी?

A)
0

B)
1

C)
1/2

D)
1/4


Correct Answer :

1


Explanation :

When coin is tossed, result will be either heads or tails.there

Total number of possible outcomes i.e. sample space n(S) : {H,T} = 2
event of getting outcomes either heads or tails n(E): {H,T} = 2

So number of probable outcomes and total number of outcomes are same i.e. 2.

So, probability of getting either heads or tails when a coin is tossed once
=> Number of favourable outcomes/Total number of possible outcomes
=> n(E)/n(S) = 2/2 = 1

Note:
Probability of getting heads when a coin is tossed once = 1/2
Probability of getting tails when a coin is tossed once = 1/2

In a box there are 4 black, 4 orange and 2 blue marbles. Two marbles are drawn at random. What is the probability that at least one marble is of blue colour?


एक बॉक्स में 4 काले, 4 नारंगी और 2 नीले कंचे है । दो कंचों को अनियमति रुप से निकाला जाता है । कम से कम एक कंचे के नीले रंग के होने की संभावना क्या है ?

A)
4/5

B)
3/5

C)
1/5

D)
None of these


Correct Answer :

None of these


Explanation :

total numbers of marbles in box= 4 black+4 orange+2 blue=10 marbles
sample space n(S)=number of ways of drawing two marbles out of 10 marbles from box= 10C2 =10!/2!*8! =10*9/2*1 =45
event of picking two marble those are not blue from out of (4 black+4 orange) 8 marbles, n(E)=8C2=8!/2!*6! =8*7/2*1=28

hence probability of drawn two marble from box which not blue p(E)=n(E)/n(S)=28/45
So probability to drown atleast one marble is blue=1-28/45=(45-28)/45=37/45

An box contains 9 red, 7 white and 4 black balls. A ball is drawn at random. What is the probability that the ball drawn is not red ?

A)
1/11

B)
9/20

C)
2/11

D)
11/20


Correct Answer :11/20
Explanation :

total numbers of balls in bucket =9 red + 7 white + 4 black=20
sample space n(S)=number of ways of drawing one ball out of 20=20C1=>20
event of drawn one red ball out of 9 red balls n(E)=9C1=>9

Hence probability p(E)=n(E)/n(S)=9/20
And probability that the ball drawn is not red , P(not-red) = (1 - 9/20) = 11/20

A number is chosen at random among the first 120 natural numbers. The probability of the number chosen being a multiple of 5 or 15 is ?

A)
1/5

B)
1/8

C)
1/6

D)
1/12


Correct Answer :1/5
Explanation :

sample space n(S)=number of ways chosen at random among the first 120 natural numbers=>120
event space n(E) =number chosen being a multiple of 5 or 15 i,e., (5, 10, ...,25, 30, ..50..,100, 105, 110, 115, 120) =24

Hence probability p(E)=n(E)/n(S)=24/120 =1/5

A single letter is selected at random from the word 'PROBABILITY' . The probability that it is a vowel, is ?

A)
3/11

B)
4/11

C)
2/11

D)
1/11


Correct Answer :4/11
Explanation :

sample space n(S)=Total number of letters=>11
event space n(E) =number of vowels i,e., (O,B,I,I) =4

Hence probability p(E)=n(E)/n(S)=4/11

If a number is chosen from the set {1, 2, 3…90}, then the probability that the chosen number is multiple of 7 is

A)
1/6

B)
2/15

C)
1/12

D)
7/30


Correct Answer :

2/15


Explanation :

All possible outcome If a number is chosen from the set {1, 2, 3…90} =sample space is n(S)=90 ways
event of getting number is multiple of 7 is n(E) ={7,14,21.....70,77,84}=12 outcomes

Therefore Probability P(E)=n(E)/n(S)=12/90=2/15

A number is randomly selected from set A ={1,2,3,} and then second number is randomly selected from set B={1,3,9} The probability that the sum of the two numbers selected will be less than or equal to 4, is:

A)
1/3

B)
5/9

C)
2/3

D)
7/9


Correct Answer :

1/3


Explanation :

All possible outcome to select the number from both set and when do sum =sample space is n(S)={1+1,1+3,1+9,2+1,2+3,2+9,3+1,3+3,3+9}=9 ways
event of getting sum of the two numbers selected will be less than or equal to 4,n(E) ={1+1,1+3,2+1}=3 outcomes

Therefore Probability P(E)=n(E)/n(S)=3/9=1/3

Which of the following has the maximum probability to be the sum of the number on top face of two dice, when both the dice thrown together?

A)
7

B)
8

C)
9

D)
10


Correct Answer :

7


Explanation :

Outcomes for Two Dice

1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

the maximum probability to be the sum of the number on top face of two dice = maximum Outcomes for that number

7 =(6, 1) , (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)
8 =(6, 2) , (5, 3), (4, 4), (3, 5), (2, 6)
9 =(6, 3) , (5, 4), (4, 5), (3, 6)
10 =(6, 4) , (5, 5), (4, 6)

Hence ,maximum probability that the sum of the two dice is seven

Two unbiased coins are tossed. What is probability of getting at most one tail ?

A)
5/4

B)
3/4

C)
1

D)
1/2


Correct Answer :

3/4


Explanation :

Sample space , n(S) = {HH, TT, TH, HT} = 4
Let Event of getting at most one tail,n(E) = {TT, TH, HT} = 3

Probability P(E) = n(E)/n(S) = 3 / 4

Three unbiased coins are tossed, what is the probability of getting at least 2 tails ?

A)
1/3

B)
1/4

C)
1/2

D)
3/2


Correct Answer :

1/2


Explanation :

Sample space ,n(S) = {TTT, HHH, TTH, THT, HTT, THH, HTH, HHT} = 8
Let Event of getting at least 2 tails,n(E) = {TTH, THT, HTT, TTT} = 4

Probability P(E) = n(E)/n(S) = 4 / 8
=> 1 / 2

In a throw of dice what is the probability of getting number greater than 5 ?

A)
2/6

B)
1/2

C)
5/6

D)
1/6


Correct Answer :

1/6


Explanation :

Sample space ,n(S) = {1, 2, 3, 4, 5, 6} = 6
Let Event of getting number greater than 5,n(E) = {6} = 1

Probability P(E) = n(E)/n(S) = 1/ 6

There are 3 children in the Khanna's family and it is known that there is at least 1 girl among the 3 children. What is the probability that all 3 of them are girls?


खन्ना के परिवार में 3 बच्चे हैं और यह ज्ञात है कि वहां 3 बच्चों में से कम से कम 1 लड़की है। सभी 3 बच्चे लड़कियां हे कि संभावना है |

A)
1/7

B)
1/8

C)
1/6

D)
2/7


Correct Answer :

1/7


Explanation :

Let G=girl, B=boy.
with three children, where each could either be a girl,the total number of possible combinations {GGG,GGB,GBG,BGG,GBB,BGB,BBG,BBB} = 8
If at least one is a girl we can rule out BBB (all boy)

thus sample space is, n(S) : {GGG,GGB,GBG,BGG,GBB,BGB,BBG} = 7
out of these only one outcomes is all girl, n(E): (GGG) = 1

Therefore, probability that all will be girl = n(E)/n(S) = 1/7

Latest Updates