Latest Elitmus Aptitude Question SOLUTION: Ram has three pyramidal dice numbered as 1,2,3,4 what is the probability that on throw of dice simultaneously the sum of number appeared on it is a prine number

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(#M40039561) ELITMUS Numerical Ability question Keep an EYE

Ram has three pyramidal dice numbered as 1,2,3,4 what is the probability that on throw of dice simultaneously the sum of number appeared on it is a prine number

Asked In Elitmus Sumit (9 years ago)

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Tej Kumar  (9 years ago)

ans-22/64
Explanation:-
2,3,5,7,11 primes
for 2 no any possibility
for 3
1+1+1=3 (1 way)
for 5
1+2+2=5 (3 way)
1+1+3=5 (3 way)
for 7
1+2+4=7 (6 way)
1+3+3=7 (3 way)
1+3+3=7 (3 way)
2+2+3=7 (3 way)
for 11
3+4+4=11 (3 way)
total no of ways=64
total event occurred is 22
therefore probability = 22/64

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Ujjwal  (9 years ago)

19/64;
1 1 1- 1 way,
1 2 2- 3 ways,
1 1 3- 3 ways,
1 2 4- 6 ways,
1 3 3- 3 ways,
3 4 4- 3 ways,
19/64 Ans...

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viek kumar  (9 years ago)

rigth ans-22/64
Explanation:-
2,3,5,7,11 primes
for 2 no any possibility
for 3
1,1,1=3 (1 way)
for 5
1,2,2=5 (3 way)
1,1,3=5 (3 way)
for 7
1,2,4=7 (6 way)
1,3,3=7 (3 way)
2,2,3=7 (3 way)
for 11
3,4,4=11 (3 way)
total no of ways=64
total event occurred is 22
therefore probability = 22/64

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SUBHAJIT DHIBAR  (9 years ago)

sum of numbers appearing on the face when three pyramidal dice is thrown simultaneously are 3, 5, 7, 11.

(1,1,1);sum=3, 1 event.

(2,2,1);sum=5, 3 events.
(3,1,1);sum=5, 3 events.

(5,1,1);sum=7, 3 events.
(4,2,1);sum=7, 3 events.
(3,3,1);sum=7, 3 events.

(4,4,3);sum=11, 3 events.

so,total number of favourable events=19.
total number of events=4^3=64.

therefore , probability=19/64.

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vinay  (9 years ago)

1/4....
At 1--> 2+3+4= 9 not prime
At 2 - - > 1+3+4=8 not prime
At 3 - - > 1+2+4= 7 is PRIME
At 4, - - > 1+2+3= 6 not prime
So probability 1/4


According to options 16/64 as far as I remember

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Vijay Pareek  (9 years ago)

for 7 12 chances for all prime numbrs 10 chances so 22/64

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GULSHAN KUMAR SINGH  (9 years ago)

22/64 or 11/32 is the correct answer

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Bala Krishna Sai  (9 years ago)

(1,1,1)- 1
(1,2,4)- 31!=6
(3,3,1)- 3
(4,4,3)- 3
(2,2,1)- 3
(1,1,3)- 3
Total- 19
Therefore, Probability= 19/(4)^3 = 19/64

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vivek kumar  (9 years ago)

(1,1,1)=1 way
(1,2,2)=3 ways
(1,3,3)=3 ways
(1,2,4)=6 ways
so probability=13/64

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Himanshu Srivastava  (9 years ago)

ans 22/64

111=3(1 way)
122=5(3 ways)
113=5(3 ways)
124=7(6 ways)
133=7(3 ways)
223=7(3 ways)
443=11(3ways)
p=22/64

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appu  (9 years ago)

3!+3!/2+3!/2=12

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nidhi  (9 years ago)

ans-4/16=1/4

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Subha Banerjee  (9 years ago)

@SUBHAJIT DHIBAR 7 is also done by (3,2,2)which has 3 ways...is it wright or wrong???plz me me

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manjeet kumar  (9 years ago)

right ans is 19/64.

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puja kumari  (8 years ago)

for 3
1+1+1=3(1 way)
for 5
1+2+2=5(3 ways)
1+1+3=5(3 ways)
for 7
1+1+5=7(3 ways)
1+2+4=7(6 ways)
1+3+3=7(3 ways)
2+2+3=7(3 ways)
for 11
3+4+4=11(3ways)
no. of ways for prime no.(E)=1+3+3+3+6+3+3+3=25
total no. of ways(S)=4*4*4=64
Therefore, probability=E/S=25/64

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pushpender  (9 years ago)

ans is 19/4

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noopur  (9 years ago)

4 is not a prme no guyz

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