Probability- Aptitude Questions and Answers

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)
T_n=2a+(n-1)d , a=5 d=5
120=4+(n−1)5
n=24
Hence probability p(E)=n(E)/n(S)=24/120 =1/5

Total number of days in a leap year = 366
It will contain 52 full weeks 2 extra days ( 52* 7 days in week + 2 days= 366 days)
Sample space for 2 extra days 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

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

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

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

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

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

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

Sample space n(S)=Total number of letters=>11
Event space n(E) =number of vowels in 'PROBABILITY' ie. (O,B,I,I) =4

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

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

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

Outcomes for Two Dice

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

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

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

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

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

There are a total of 52 cards.
Now, there are 26 red cards and 12 face cards, 6 of which are red (double counted)

So, the probability of having a card's red card or face card
P(Red u King)=P(Red)+P(face)-P(Red n face)
=26/52 + 12/52 - 6/52
=32/52
=8/13