Latest TCS Aptitude Question SOLUTION: there r three dices one blue and two red.find the probability that the no appered in red dices is more than the blue dice?

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there r three dices one blue and two red.find the probability that the no appered in red dices is more than the blue dice?

Asked In TCS sagar kumar (8 years ago)

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prasad gaikwad  (8 years ago)

if blue die shows 1, red die can have any of the 5 nos(2, 3, 4, 5, 6) which are greater than 1 hence no of ways = 5C1*5C1
blue die= 2, red die= any one from (3,4,5,6) hence= 4C1*4C1
blue die=3, red die= 3C1*3C1
blue die=4, red die = 2C1*2C1
blue die=5, red die=1C1*1C1
total possibilities as per given condition= 25+16+9+4+1=55
hence probability= 55/216

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anand jha  (8 years ago)

We have 3 dices it means total probability is 216. Now the no. on red dices should be more than the blue.
then the no. on red which are less or equal to blue are..
(1,1).....(1,5)
(2,1).....(2,4)
(3,1).....(3,3)
(4,1).....(4,2)
(5,1)
total 15
now total outcomes are 216
therefore favourable is 216-15=201
hence probability is 201/216=67/72 ans.

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Abhi  (8 years ago)

Total possibility - 216
No. Appeared in red dices more than blue dices- 55
Probability - 55/216

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Biswadeep Sen  (8 years ago)

considering the case where n(blue)>n(red) we will have 4/18 as 3,4,5 and 6 are the available condition , so the solution should me 1-4/18=14/18

Is this solution Helpfull? Yes (4) | No (6)  

monalika  (8 years ago)

there are 3 dices.....so total no of cases will be (6*6*6)
Now if blue dice will have 1,rd dice will have nos more than 1,similarly if blue dice will have 2,red can have 4 nos greater than 2 and in the similar way goes on.So,it will continue upto 5 for blue dice,then red dice will have 6 which is greater than 5.........so no of cases for blue dice having no smaller than red one=(1+2+3+4+5)ie 15.........so (216-15)ie 201
therefore,prob= 201/216 ie 67/72

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vikash rawat  (8 years ago)

55/216...is the correct answer...

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Anjali Sadhukhan  (8 years ago)

While throwing one red dice along one blue dice, we can find 15 ways where red dice numbrs are grater than the blue, hence 15/216. Please correct me if I have mistaken smwhr

Is this solution Helpfull? Yes (1) | No (5)  

praveen  (8 years ago)

please tell the correct answer.I dont know the probability

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richa sharma  (8 years ago)

correct ans/////???????

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shahbaz khan  (8 years ago)

solve it by bottom-up approach
first,find the number of ways by which the number on blue dice can be greater than the numbers on red dice
Blue Red1 Red2 Total_no_of_ways
1 - - 0
2 1 1 1
3 1 1 ......
3 1 2 4
3 2 1
3 2 2
4 .... ..... 9
5 ..... ..... 16
6 ..... ..... 25

total no of ways=(1+4+9+16+25)=55+no_of ways in which all numbers can beequal(=6)=61
total no ways in which numbers can appear on all 3 dices=6*6*6=216
probability that numbers on blue dice is greater than or equal to the numbers on red dices=61/216
Required probabilty=1-61/216=155/216

Is this solution Helpfull? Yes | No (2)  

bhumika  (4 years ago)

if blue die shows 1, red die can have any of the 5 nos(2, 3, 4, 5, 6) which are greater than 1 hence no of ways = 5C1*5C1
if blue die shows 2, red die can have any of the 4 nos(3, 4, 5, 6) which are greater than 1 hence no of ways = 4C1*4C1
if blue die shows 3, red die can have any of the 3 nos(4, 5, 6) which are greater than 1 hence no of ways = 3C1*3C1
if blue die shows 4, red die can have any of the 2 nos(5, 6) which are greater than 1 hence no of ways = 4C1*4C1
if blue die shows 5, red die can have any of the 1 nos( 6) which are greater than 1 hence no of ways 1C1*1C1
total possibilities as per given condition= 25+16+9+4+1=55
hence probability= 55/6^3

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