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what is the probability of getting a sum of 5 before 7 when 3 dice are thrown?
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- sry friends its 2/5
- 11 years agoHelpfull: Yes(10) No(11)
- answer is 1/36
- 11 years agoHelpfull: Yes(7) No(8)
- 2/7 is ans
- 11 years agoHelpfull: Yes(5) No(0)
- 1/18 ans. total ways of getting 5 is 12 and total number of outcomes is 216(6*6*6) hence 1/18
- 11 years agoHelpfull: Yes(3) No(4)
- ans is 1/54 i.e probability of getting 5 to total probability
- 11 years agoHelpfull: Yes(2) No(4)
- @Shilpi : how did u get that 21 times..plz explain ..
- 11 years agoHelpfull: Yes(2) No(1)
- 1/18 ans. total ways of getting 5 is 12 and total number of outcomes is 216(6*6*6) hence 1/18
- 11 years agoHelpfull: Yes(1) No(5)
- 7/72(5 comes before 7, 21 times and total outcomes are 216)
- 11 years agoHelpfull: Yes(1) No(1)
- plz explain
- 11 years agoHelpfull: Yes(1) No(1)
- ans is 6/11
- 11 years agoHelpfull: Yes(0) No(10)
- p(5)=1+1+3 =3!/2!=3 ways
p(5)=1+2+2 =3!/2!= 3 ways , total p(5)=3+3=6 ways
p(7)= (1+1+5)=3!/2!=3 ways
p(7)= (1+2+4)=3!=6 ways
p(7)= (1+3+3)=3!/2!=3 ways
p(7)= (2+2+3)=3!/2!=3 ways
total p(7)=3+6+3+3 =15
probability(5 before 7)= p(5)/{p(5)+p(7)} = 6/(6+15) = 6/21 = 2/7 ans
- 8 years agoHelpfull: Yes(0) No(0)
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