4. A dice is thrown 10 times. The number 6 appears
3 times. What is the probability of getting number 6?
Answers
ANSWER
From given, we have,
Sum 3:(1,1,1) ==> Contributing only 1 distinct triplet.
Sum 4: (1,1,2) ==> Contributing 3 distinct triplets
Sum 5: (1,2,2)and (1,1,3) ==> Contributing 6 distinct triplets
Sum 6: (1,1,4),(1,2,3) and (2,2,2) => Together contributing 10distinct triplets
Sum 7: (1,1,5),(1,2,4),(1,3,3)and (2,2,3) => Together contributing 15 distinct triplets.
Sum 8: (1,1,6),(1,2,5),(1,3,4),(2,3,3) and(2,4,2)==> Together contributing 21 distinct triplets.
Sum 9: (1,2,6),(1,3,5),(1,4,4),(2,3,4),(2,5,2) and (3,3,3) ==> Together contributing 25 distinct triplets.
Sum 10: (1,3,6),(1,4,5),(2,2,6),(2,3,5),(2,4,4) and (3,3,4)==> Together contributing 27 distinct triplets.
Therefore number of favorable cases =1+3+6+10+15+21+25+27=108.
Therefore, probability =
216
108
=
2
1