Elitmus
Exam
Numerical Ability
Probability
a two dices have four faces (1, 2, 3, 4) both are simultaneously threw what is the probability of sum of the faces is prime
Read Solution (Total 6)
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- 9/16
Total outcomes with four faces (1,2,3,4) = 4x4 = 16
Favorable cases are (1,1) (1,2) (1,4) (2,1) (2,3) (3,2) (3,4) (4,1) (4,3), total 9
So required probability = 9/16 - 7 years agoHelpfull: Yes(24) No(1)
- We have four faces..1,2,3,4..
Now the favourable cases are
1,1 2,1 3,2 4,1
1,2 2,3 3,4 4,3
1,4
Now total favourable 9 cases and
Total are 16
Hence the ans is 9/16 - 7 years agoHelpfull: Yes(1) No(1)
- 7/16 incorrect
9/16 currect bkz missing (1,4)(4,1) it is the sum of the 5 which is also prime number - 7 years agoHelpfull: Yes(1) No(0)
- 9/16
2 dices with 4 faces so 4*4=16
total possible of getting prime num is 9
so probability is 9/16 - 6 years agoHelpfull: Yes(1) No(0)
- total = 4*4=16 is sample space for throwing two dices. now using that four combination sum of primes comes only (1,1) or (2,1) or (1,2 ) or (3,4) or (4,3) or (2,3) or (3,2) these are all possible combination .
so answer is 7/16. - 7 years agoHelpfull: Yes(0) No(4)
- 9/16 ans
total cases= 4*4=16
favourable output prime number sum{(1,1),(1,2), (2,1), (2,3), (3,2), (1,4), (4,1), (3,4), (4,3) } ie total=9
so probability=9/16 ans - 7 years agoHelpfull: Yes(0) No(0)
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