Latest TCS Aptitude Question SOLUTION: There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement. What is the probability atleast three are red!

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There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement. What is the probability atleast three are red!

Asked In TCS Karthika (10 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

suraj prasad  (10 years ago)

5 ball out of 21 ball can be selected in 21C5 ways.
at least means favorable events may be (6c3*15c2 + 6c4*15c1 + 6c5)/21c5

Is this solution Helpfull? Yes (45) | No (20)  

Arshal Jain  (10 years ago)

case 1 : 3 red balls : 6x6x6x15x15
case 2 : 4 red balls : 6x6x6x6x15
case 3 : 5 red balls : 6x6x6x6x6
favourable cases = cases ( 1 + 2 + 3) = 30456
total cases = 21x21x21x21x21 = 21^5
probability = 376/50421

Is this solution Helpfull? Yes (44) | No (10)  

Karthika  (10 years ago)

@Suraj,chirag : Consider replacement..
and @Arshal: the method yu hav done is right..but the final answer is 936/50421=312/16807.

Is this solution Helpfull? Yes (13) | No (2)  

harsh  (9 years ago)

At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement.
case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21
case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21
case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21

Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)
= 312/16807

Is this solution Helpfull? Yes (11) | No   

chirag jain  (10 years ago)

We have to select 5balls so and atleast 3should be red..
so more than 3could also be red..
total ways so selecting 5balls in such a way = 6c3*15c2+ 6c4*15c1+ 6c5

And the total no of ways for selecting 5balls is 21c5
so the answer will be 37/323

Is this solution Helpfull? Yes (6) | No (7)  

Amsaveni Natarajan  (10 years ago)

((6c3*15c2)+(6c4*15c1)+(6c5))/21c5

Is this solution Helpfull? Yes (2) | No (5)  

Arshal Jain  (10 years ago)

yeah you are right ....i multiplied wrong terms in hurry

Is this solution Helpfull? Yes (1) | No (2)  

S.B.Yasaswini  (10 years ago)

((6c3*15c2)+(6c4*15c1)+(6c5))/21c5

Is this solution Helpfull? Yes (1) | No   

NITESH RANJAN  (10 years ago)

pow(5,7)*5

Is this solution Helpfull? Yes | No (13)  

Karthika  (10 years ago)

@Arshal Jain : This is the right answer!!

Is this solution Helpfull? Yes | No   

Sushil baran Mahato  (10 years ago)

((6c3*15c2)+ (6c4*15c1)+ 6c5)/21c5=2331/20349=259/2261

Is this solution Helpfull? Yes | No (1)  

Bhrugu  (8 years ago)

As out of 6 red 8 blue and 7 green balls are there we have space = 21
event is ,
case 1= 6c5=6( all are red)
case 2=6c3+8c2=48(3 red and 2 blue)
case 3= 6c3+7c2=41(3 red and 2 green)
hence event=6+48+41=95
so , n(E)/n(s)=95/21

ANSWER= 95/21

Is this solution Helpfull? Yes | No (3)  

shashank gupta  (6 years ago)

total ways so selecting 5balls in such a way = 6c3*15c2+ 6c4*15c1+ 6c5


And the total no of ways for selecting 5balls is 21c5
so the answer will be 37/323 ""

Is this solution Helpfull? Yes | No   

sumit  (5 years ago)

70-54=16 i.e square of 4.
54-45=9=i.e square of 3
45-41=4 i.e square of 2
4,3,2 are following a pttern in decreasing order so
41-x=1 i.e square of 1.

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