Latest TCS Aptitude Question SOLUTION: 3 dies are thrown. What is the probability that sum ten appears??

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3 dies are thrown. What is the probability that sum ten appears??

Asked In TCS haran (10 years ago)

Solved S.SHAFI Read Solution (10)

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Read Solution (10) :

S.SHAFI  (10 years ago)

When 3 dice are thrown, total number of possibilities are n(s) = 6^3 = 216

Sum is 10 : (1,3,6),(1,4,5),(1,5,4),(1,6,3)
(2,2,6),(2,3,5),(2,4,4),(2,5,3),(2,6,2)
(3,1,6),(3,2,5),(3,3,4),(3,4,3),(3,5,2),(3,1,6)
(4,1,5),(4,2,4),(4,3,3),(4,4,2),(4,5,1)
(5,1,4),(5,2,3),(5,3,2),(5,4,1)
(6,1,3),(6,2,2),(6,3,1)
Probability of getting a sum of ten is 27/216 = 1/8

Is this solution Helpfull? Yes (42) | No (3)  

SUNNY KUMAR SHAW  (10 years ago)

IF THE FIRST DICE HAS 1 THEN THE REST SHOULD GIVE A SUM OF 9 :- (3,6) (4,5) (5,4) (6,3)= 4 SAMPLES
IF THE FIRST DICE IS 2 THEN THE REST SHOULD GIVE A SUM OF 8:- (2,6) (3,5) (4,4) (5,3) (6,2)= 5 SAMPLES
IF IT IS 3 THEN A SUM OF 7 :- 6 SAMPLES
IF 4 THE A SUM OF 6:- 5 SAMPLES
IF 5 THEN A SUM OF 5:- 4 SAMPLES
IF 6 THEN A SUM OF 4:- 3 SAMPLES

THEREFORE DESIRED SAMPLE IS= 27
TOTAL SAMPLE IS 216
THEREFORE PROBABILITY IS:(27/216) OR 1/8

Is this solution Helpfull? Yes (12) | No   

Mallela Srilakshmi  (10 years ago)

we will get 27 possibilities when we throw 3 dice to get sum 10. they are
{(136)(145)(154)(163)(226)(235)(244)(253)(262)(316)(325)(334)(343)(352(361)(415)(424)(433)(442)(451)(514)(523)(532)(541)(613)(622)(631).
so probability is 27/216=1/8=0.125

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Sarmistha Sarkar  (10 years ago)

n(s)=6^3=216
sum=10
possible:{(6,3,1),(6,2,2),(541),(5,3,2),(4,3,3),(4,4,2)}
(6,3,1):3!=6
(6,2,2)=3
(541)=3!=6
(5,3,2)=3!=6
(4,3,3)=3
(4,4,2)=3
so total=27
hence answer 27/216=1/8

Is this solution Helpfull? Yes (3) | No   

Santosh kumar sahu  (9 years ago)

short cut :- By 3 dies-

number sum

3 1
4 3
5 6
6 10
7 15
8 21
9 25
10 27
11 27
12 25
13 21
14 15
15 10
16 6
17 3
18 1


sum of 10 =27

than 27/6*6*6=1/8

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nupur kamboj  (10 years ago)

1/108 bcz for sum to be 10 there are two possibilities i.e. (6,4) and (4,6).sample space is 256.and the probability comes out to be2/256.

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Mantripragada Nikhil  (10 years ago)

(1,3,6)(2,3,5)(2,2,6)(2,4,6)(3,3,4)(4,1,5)(4,2,4)...this outcomes may be interchanged 3 times each...so no.of favourable events are 3*7=21 chances wid total outcomes as 6^3=216...so probability=21/216=7/72.

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RISHI SHEKHAR JHA  (10 years ago)

When 3 dice are thrown, total number of possibilities are n(s) = 6^3 = 216
possibilities that the sum appears 10 are:
334,352,361
451,442,443
while 334 can be arranged in 3p2 ways,352 can be arranged in 3p1 ways....so on
so adding all these possible arrangements ,we get the sum total of 27
hence 27/216=1/8

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Prasoon Singh  (9 years ago)

To be able to roll a 10 with 3 dice, the first 2 dice must give a value between 4 and 9 (probability of 3/4), in which case there is a 1/6 chance that the third die will make the total 10. So, the overall probability is 3/4*1/6=1/8

This does assume some knowledge of the 2-dice distribution, which has the probability ratios 1:2:3:4:5:6:5:4:3:2:1 for 2 through 12. So, the total weight, out of 36, for 4 through 9, is 3+4+5+6+5+4 = 27.

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Pratik sharma  (9 years ago)

Total case is 216 i.e. 6^3
Now
(1,3,6)->3!=6
(3,3,4)->3
(4,4,2)->3
(6,2,2)->3
(5,3,2)->3!=6
(5,4,1)->3!=6

thus 6+3+3+3+6+6=27
so probability is 27/216=1/8

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