Latest Wipro Aptitude Question SOLUTION: There are 3 red balls,3 green balls and 3 blue balls are present.what is the probability of atleast two of them are of same colour if 3 balls are drawn?

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There are 3 red balls,3 green balls and 3 blue balls are present.what is the probability of atleast two of them are of same colour if 3 balls are drawn?

Asked In Wipro KOMAL KUMARI (10 years ago)

Unsolved Read Solution (26)

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Read Solution (26) :

Girjesh kumar  (10 years ago)

Total combinations = 9c3 = 84
if we calculate the combinations in which three balls are of different colours,
then combinations will be = 3c1*3c1*3c1 = 27
so final probability of atleast two of them are of same colour if 3 balls are drawn = 1 - (27/84)
= 57/84

Is this solution Helpfull? Yes (31) | No (5)  

S.Abirami  (9 years ago)

3 balls are drawn two of them are of same color:
there are three cases (i)red (ii)green (iii)blue
(i)red means 2 of them are red and any one out of 6 other balls=3c2*6c1
(ii)green means 2 of them are green and any one out of 6 other balls=3c2*6c1
(iii)blue means 2 of them are blue and any one out of 6 other balls=3c2*6c1
initially 3 balls are drawn out of 9 balls=9c3=84
therefore total probability is (3c2*6c1+3c2*6c1+3c2*6c1)/84=54/84=27/42 is the answer.

Is this solution Helpfull? Yes (31) | No (19)  

sai  (9 years ago)

atleast 2 of same colour.so there can be 3 of same colour also.
total probability : 9c3=84
probabilty that 2 of same colour : 3c2+6c1=18 .there can be 3 such cases so 3(3c2+6c1)= 53
probablity that 3 of same colour = 3c3=1 and three such cases of 3
now = 54+3/84 =57/84

Is this solution Helpfull? Yes (16) | No (1)  

chaitanya  (9 years ago)

3 balls are drawn two of them are of same color:
there are three cases (i)red (ii)green (iii)blue
(i)red means 2 of them are red and any one out of green or blue = (3c2*3c1*3c0)+(3c2*3c0*3c1)
(ii)green means 2 of them are green and any one outof blue or red =(3c1*3c2*3c0)+(3c0*3c2*3c1)
(iii)blue means 2 of them are blue and any one out of green or red=(3c0*3c1*3c2)+(3c1*3c0*3c2)
in the Q it is given atleast two...it means 2 and more than 2 ==>3
then for all the 3 of same clour is =(3c3*3c0*3c0)+(3c0*3c3*3c0)+(3c0*3c0*3c3)
initially 3 balls are drawn out of 9 balls=9c3=84
prbability =((3c2*3c1*3c0)*3)+3)/84 = 57/84

Is this solution Helpfull? Yes (7) | No   

vibhaw upadhyay  (9 years ago)

prob of at least 2 same ball=(1-prob of all diff ball)
=1-{(3C1*3C1*3C1)/9C3}
=1-{(3*3*3)/(9!/3!*6!)}
=1-{(9/84)}
=75/84

Is this solution Helpfull? Yes (6) | No   

BIKKY GUPTA  (9 years ago)

3 balls are drawn two of them are of same color:
there are three cases (i)red (ii)green (iii)blue
(i)red means 2 of them are red and any one out of green or blue = (3c2*3c1*3c0)+(3c2*3c0*3c1)
(ii)green means 2 of them are green and any one outof blue or red =(3c1*3c2*3c0)+(3c0*3c2*3c1)
(iii)blue means 2 of them are blue and any one out of green or red=(3c0*3c1*3c2)+(3c1*3c0*3c2)
in the Q it is given atleast two...it means 2 and more than 2 ==>3
then for all the 3 of same clour is =(3c3*3c0*3c0)+(3c0*3c3*3c0)+(3c0*3c0*3c3)
initially 3 balls are drawn out of 9 balls=9c3=84
prbability =((3c2*3c1*3c0)*3)+3)/84 = 57/84

Is this solution Helpfull? Yes (3) | No   

Shaik.Abdul Nawaz   (10 years ago)

i think ans will b 1/4

Is this solution Helpfull? Yes (2) | No (4)  

chaitanya  (9 years ago)

3 balls are drawn two of them are of same color:
there are three cases (i)red (ii)green (iii)blue
(i)red means 2 of them are red and any one out of green or blue = (3c2*3c1*3c0)+(3c2*3c0*3c1)
(ii)green means 2 of them are green and any one outof blue or red =(3c1*3c2*3c0)+(3c0*3c2*3c1)
(iii)blue means 2 of them are blue and any one out of green or red=(3c0*3c1*3c2)+(3c1*3c0*3c2)
in the Q it is given atleast two...it means 2 and more than 2 ==>3
then for all the 3 of same clour is =(3c3*3c0*3c0)+(3c0*3c3*3c0)+(3c0*3c0*3c3)
initially 3 balls are drawn out of 9 balls=9c3=84
prbability =((3c2*3c1*3c0)*6)+3)/84 = 57/84

Is this solution Helpfull? Yes (2) | No   

Dr.Dipin Singh  (10 years ago)

Total combinations = 9C3=84
only one combination is undesirable when three balls are from 3 different colours.
so
probability of atleast two of them are of same colour if 3 balls are drawn= 83/84

Is this solution Helpfull? Yes (1) | No (9)  

Dr.Dipin Singh  (10 years ago)

only one combination is undesirable when three balls are from 3 different colours.

3 combinations are there when 3 balls are of same (one) colour.
6 combinations are there when balls are of 2 colours.
so total possible combinations =10
so
probability of atleast two of them are of same colour if 3 balls are drawn= 9/10

Is this solution Helpfull? Yes (1) | No (9)  

Saumya  (9 years ago)

N(S)= 9C2
N(E)= 21
(SINCE, the combination may be (RRG,GRR,RGR,GGR,GRG,RGG,RRB,BRR.RBR,GGB,BGG,GBG,BBG,BGB,GBB,BBR,BRB,RBB,RRR,BBB,GGG)
HENCE, P(E)=N(E)/NS)
i.e. 21/9C3 = 1/4

Is this solution Helpfull? Yes (1) | No   

p.arun  (10 years ago)

total balls 9c3=84
atleast 3drawn form each balls=(3c3*3c2)+(3c3)=10


n(E)/n(s)=84/10=42/5

Is this solution Helpfull? Yes | No (10)  

Ann Theressa  (10 years ago)

ans: 53/56

The probability that atleast two of them are of same colour = 1 - probability that 3 balls drawn are different colour

probability that 3 balls drawn are different colour = 3/9 * 3/8 * 3/7
= 3/56

Therefore
The probability that atleast two of them are of same colour = 1-3/56
= 53/56

Is this solution Helpfull? Yes | No (7)  

vijayaramireddy  (10 years ago)

ans:-27

we have 3g,3b,3r according given condition atleast two of them are of same colour if 3 balls are drawn

3c2*3c1*3c0+3c1*3c2*3c0+3c0*3c1*3c2

9+9+9=27

Is this solution Helpfull? Yes | No (3)  

Shaik.Abdul Nawaz   (10 years ago)

answer is 25/28 bcoz total combination is 9c3
atleast 2 balls having same colour either red nd green is 3c23c1 r green nd red is 3c13c2 like wise green nd blue is 3c23c1 r blue nd green 3c13c2 likewise blue nd red is 3c23c1 r red nd blue is 3c13c2 then add al we get total is 72 r three balls r having same colour either red r green r blue is 3c3+3c3+3c3 =3
72+3=75/9c3=25/28

Is this solution Helpfull? Yes | No (4)  

anchal jain  (10 years ago)

2/3 because total outcomes will be 9 nd favourable will be 6

Is this solution Helpfull? Yes | No (2)  

dheeraj  (9 years ago)

atleast two of them are of same colour=3(3c2*9c1/9c3)=42/84
all of them are of same colour=3(3c3/9c3)=3/84

Is this solution Helpfull? Yes | No (3)  

santhi   (9 years ago)

3*3c2/9c2=1/4

Is this solution Helpfull? Yes | No (3)  

ruchi  (9 years ago)

27/42
drawn two of them are of same color:
there are three cases (i)red (ii)green (iii)blue
(i)red means 2 of them are red and any one out of 6 other balls=3c2*6c1
(ii)green means 2 of them are green and any one out of 6 other balls=3c2*6c1
(iii)blue means 2 of them are blue and any one out of 6 other balls=3c2*6c1
initially 3 balls are drawn out of 9 balls=9c3=84

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manish kumar  (9 years ago)

6(3c2.3c1)+3.3c3/9c3=57/84=19/28

Is this solution Helpfull? Yes | No (2)  

Chitransh Teotia  (9 years ago)

Atleast 2 of them must be of same color
so,we have to find out the probability for 2 or more balls to be of same color.
so,for 2 or more red balls to be of same color the combinations are ->3c2*7c1.
similarly,for blue and green balls combinations are the same.
Also, the total number of balls fetched is 3 out of nine so combination is->9c3
so, the probability must be {3(3c2*7c1)}/9c3
which will come out to be 63/84.
THIS IS THE CORRECT ANSWER

Is this solution Helpfull? Yes | No (2)  

vineet  (9 years ago)

((3C2*6C1)+3C3)/9C3

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Jyothi Prakash Thaduri  (9 years ago)

((3c2*7c1)*3)/9c3=1/4

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vijay karthik  (9 years ago)

ans;9/28
3 ball drown pro.=84
3 same color=3times
atleast 2 same color (ex take RGB == red comes in 2 times RRG RGR GRR RRB RBR BRR so 6 times) 6*3=18
18+3=21
ans 21/84===1/4

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satish  (8 years ago)

correct ans is 5/14 n(s)= 9c3= 84 n(e)= 9*3+3= 30 p(e)=30/84=5/14

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Biswadip Mandal  (7 years ago)

total combination = 9c3 = 84;
when 2 of the colors are same:
choose 1 color : 3c1
choose any 2 balls of that particular color: 3c2
when 3 colors are same:
choose any color: 3c1
total required combination = 3c1*3c2+3c1=12
probability = 12/84 = 1/7

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