Latest TCS Aptitude Question SOLUTION: Eri threw two dice and his score is the sum of the values shown .Sandy throws one die and her score is the square of the value shown. find the probability that sandy's score wil be

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Eri threw two dice and his score is the sum of the values shown .Sandy throws one die and her score is the square of the value shown. find the probability that sandy's score wil be strictly greater than eri's score?

Asked In TCS Karthika (10 years ago)

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Shashi Kumar Gupta  (10 years ago)

the maximum sum of two dice is 12 i.e Eri's maximum value is 12.
now,if Sandy get more than 12 then her value will be strictally greater than Eri.
so, the only square of 4,5,6, is greater than 12..
probability=3/6=>1/2

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Bholanath   (10 years ago)

Sandy may score 1,4,9,16,25 or 36 that is square of all the possibilities of one dice thrown by sandy.
Eri's total possibilities = 36 for two dice
sum of two dice may be 2,3........,12

according to question Sandy's score strictly greater than Eri's so

for Sandy's 1 there is no any possibilities to get greater
for 4 Eri's possiblities {(1,1),(1,2),(2,1)}=3
for 9 Eri's possiblities i.e. sum less than 9= {(1,1),.....(1,6),(2,1)......(2,6),(3,1)....(3,5),(4,1)....(4,4),(5,1)....(5,3),(6,1),(6,2)}=26
for 16 Eri's possibilities all combination (because maximum sum=12 of two dice)=36
for 25 Eri's possibilities =36
for 36 Eri's possibilities= 36

so total no of favorable cases=3+26+36+36+36=137
and total no. of sample space=6*6*6=216
so,
probablitiy=137/216

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Kalidindi Mounika  (10 years ago)

possibilities for sandy's score greater than Eric is {(4,4),(5,5),(6,6)}
so probability will be 3/36=1/12

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muttareddy veera mani devi  (10 years ago)

if sandy score is 1-1^2=1 eri score=0
if 2-2^2=4 =3
if 3-3^3=9 =17
if 4-4^2=16,5^2=25,6^2=36 eri score may be 3*36=108
so total values=120
the probability may be 1/120

Is this solution Helpfull? Yes (1) | No (4)  

prarit jain  (10 years ago)

as there is only one probability ie (1,1) den the sandy score is 4 so probability is 1/36.

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GANESH RAVVA  (10 years ago)

eri's(x+y) sandy(z^2)
1 1 5ways(i.e 2^2=4 3^2=9 4^2=16 5^2=25 6^2=36)
1 2 5
1 3 4
1 4 4
1 5 4
1 6 4
2 1 4
2 2 2
.
.
.
2 6 3ways
3 1 4
.
.
.
.so on like this..add every one...we get total of 132
ans=probability is (132/216=11/18)
(dont think it is so big procedure bacause we can simply right combinations i.e for example..2 4 is 4ways
similarly 5 1
3 3
4 2
1 5
answer will be same..so it can be written fast
if anything wrong pls post new answer and corct me.:)

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HARSH  (10 years ago)

pls send me tcs apti papers 2013.......my id-rharsh170@gmail.com

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