2 dies are thrown Probability of Getting a sum of 5 before 7 is

sum 5 events: (1,4),(2,3),(3,2),(4,1)
total probability of getting 5 is 4/36.
sum 7 events: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1).
total probability of getting 7 is 6/36.
So Probability of Getting a sum of 5 before 7 is:-
(4/36)/(4/36)+(6/36)
2/5.
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to get sum=5,we have 4 outcomes
and to get sum=7,we have 6 outcomes
and in order to get sum=5 before 7,total outcomes for sum=7 will remain 32
so ans. is (4/36)*(6/32)=0.0208
10 years agoHelpfull: Yes(2) No(1)
2/5 i think
10 years agoHelpfull: Yes(1) No(0)
(1,4)(4,1)(2,3)(3,2)so 4/36=1/9 is the answer
10 years agoHelpfull: Yes(0) No(0)
4/(36-6)=2/15
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EVENT OF GETTING SUM 5- (1,4)(4,1)(2,3)(3,2)
P(5)= 4/36= 1/9
EVENT OF GETTING SUM 5- (1,6)(6,1)(2,5)(5,2)(3,4)(4,3)
P(7)= 6/36 = 1/6
P(NOT 7)= 1-1/6 = 5/6
P(5 INTERSECTION (NOT 7))= 1/9* 5/6= 5/54
WE HAVE TO FIND P(5/NOT 7)= P(5 INTERSECTION (NOT 7))/P(NOT7)
=5/54*6/5 = 1/9 (ANS.)
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sums probability of getting
2 1
3 2
4 3
5 4
6 5
7 6
8 5
9 4
10 3
11 2
12 1
probability of getting sum of 5 are 4
before 7 1+2+3+4+5+6=21
so 4/21
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P(5) : P(7)
-------------
2/18 : 3/18

= 2 : 3
2 2
and probability of 5 before 7 is ----- = ---
2+3 5
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2 dies are thrown. Probability of Getting a sum of 5 before 7 is??