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Numerical Ability
Permutation and Combination
There are 3 red, 3 blue, 3 green marbles. What is probability that atleast two of them are of same color. If 3 marbles are done without replacement.
Read Solution (Total 13)
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- ((3c2*6c1)/9c3)*3+(3c3/9c3)*3
=19/28 - 11 years agoHelpfull: Yes(35) No(4)
- total sample=9c3
case 1: when we take 2 marbles of same color
ie, of red or green or blue and one different
3*(3c2*6c1)/9c3
case 2: when we take 3marbels of same color
3*(3c3*6c0)/9c3
add both u will get answer....... - 11 years agoHelpfull: Yes(25) No(1)
- ans is 9/14.
there are 3 cases (3c2*6c1)/9c3=3/14
all the 3 cases are same so
3*3/14=9/14. - 11 years agoHelpfull: Yes(10) No(18)
- ans is 19/28
{3*(3c2*6c1)/9c3+3*(3c3*6c0)/9c3} - 11 years agoHelpfull: Yes(8) No(2)
- Total possible outcomes 9c3
probability(atleast two of them are same color)=1-probability(3 different)
=1-(3c1*3c1*3c1)/9c3
=19/28 - 10 years agoHelpfull: Yes(7) No(0)
- ans is given as 156/120
- 11 years agoHelpfull: Yes(0) No(10)
- Ans: 3(3c2*6c1)/ (9c3)
- 11 years agoHelpfull: Yes(0) No(9)
- Ans: 3(3c2*6c1)/ (9c3)
- 11 years agoHelpfull: Yes(0) No(8)
- Ans: 3(3c2*6c1)/ (9c3)
- 11 years agoHelpfull: Yes(0) No(9)
- Ans: 3(3c2*6c1)/ (9c3)
- 11 years agoHelpfull: Yes(0) No(6)
- given it is atleast two of them are of same colour so there are 2 possibilities
1. As 3 marbles are drawn atleast two of same colour and 1 from rest of the 6=3c2*6c1
2. 3 of same colour and none from the rest 6=3c3*6c0
so (3c2*6c1+3c3*6c0)/9c3 =19/84
- 10 years agoHelpfull: Yes(0) No(2)
- possible outcome={RRB,RBR,BRR,RRG,RGR,GRR,BBR,BRB,RBB,BBG,BGB,GBB,GGR,GRG,RGG,GGB,,GBG,BGG}=18
(18*3*2*3)/(9*8*7)=9/14
ANS=9/14 - 9 years agoHelpfull: Yes(0) No(1)
- 3/8 is the the correct answer
- 7 years agoHelpfull: Yes(0) No(0)
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