Elitmus
Exam
Numerical Ability
Probability
A fair dice is tossed six times. Find the probability of getting a third six in the sixth throw
Read Solution (Total 3)
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- For the first 2 sixes (withing first 5 throws), the probability is: 5C2/6^2
For the rest throws, the no.s may appear-> 1,2,3,4,5. The probability is 5^3/6^3
Now for the 6th throw to be 6, the probability is: 1/6
So, P(Getting a 3rd six in the 6th throw)= (5C2*5^3)/6^6 (Ans). - 8 years agoHelpfull: Yes(17) No(5)
- total possible outcomes are = 6^6
now at sixth throw it is confirmed that we will get a 3, so that means we are left with two more 3s which could possibly occur in any 2 of the first five throws..
So total number of ways to do so is 5C2(we can not apply permutation here becoz that will yeild the same result)
Now 2 places among 5 are occupied by 3s so rest 3 places/throws can result in any of the numbers-{1,2,4,5,6}...so we can get these numbers in 5^3 ways.
so the required probability would be P = (5C2*5^3)/(6^6)...thats how it is. - 8 years agoHelpfull: Yes(6) No(1)
- 5c2/6*6*6*6*6*6
- 8 years agoHelpfull: Yes(1) No(1)
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