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

• 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
  
  
• 10 years agoHelpfull: Yes(42) No(3)
• 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
• 10 years agoHelpfull: Yes(12) No(0)
• 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
• 10 years agoHelpfull: Yes(3) No(0)
• 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
• 10 years agoHelpfull: Yes(3) No(0)
• 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
• 10 years agoHelpfull: Yes(2) No(0)
• 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.
• 10 years agoHelpfull: Yes(0) No(8)
• (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.
• 10 years agoHelpfull: Yes(0) No(4)
• 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
  
• 10 years agoHelpfull: Yes(0) No(0)
• 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.
• 10 years agoHelpfull: Yes(0) No(0)
• 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
• 10 years agoHelpfull: Yes(0) No(0)