JEE Probability 2
A dice is rolled 6 times. The probability that number of times prime numbers appears is more than number of times an even number appears is γ . Calculate ⌊ 1 0 0 0 γ ⌋ .
This problem is a part of My picks for JEE 1
The answer is 403.
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2 solutions
This is arguably one of the most beautiful questions that I've had the honor of solving and one of the most beautiful solutions that I've had the honor of finding myself. Looks like someone just beat me to it :/ But anyway, thanks!
Extremely elegant.
Let a roll 1 or 2 be called a (neutral event). Let a roll 3 or 5 be called b (increases the number of primes) Let a roll 4 or 6 be called c (increases the number of even numbers) Then we have to find the number of ways we can arrange the letters a, b & c such that N ( b ) > N ( c ) . To me that yields: All b's : ( 6 ⋅ 2 0 6 ) 5 b's, 1 either a or c : ( 5 ⋅ 2 1 6 ) 4 b's, 2 either a or c : ( 4 ⋅ 2 2 6 ) 3 b's, at most 2 c's : 6 \choose 3 \cdot (2^3 - 3 \choose 3) 2 b's, at most 1 c : 6 \choose 2 \cdot ( 4 \choose 1 + 4 \choose 0) 1 b, 0 c's : 6 \choose 1 \cdot ( 5 \choose 0) . Adding all those ways together and dividing my the total number of ways the letters a, b & c can be arranged in a string of length 6 yields (\ \gamma = \frac{294}{3^6)=0.3827 )
| prime | not prime | |
| even | 2 : 6 1 | 4 , 6 : 3 1 |
| not even | 3 , 5 : 3 1 | 1 : 6 1 |
Let M p the odds of having more primes than even numbers, M e the odds of having more even numbers than primes, and E q the odds of having as much primes as even numbers. We have of course M p + M e + E q = 1 .
As there's a symmetry in the odds, M p = M e , which gives M p = 2 1 − E q . So, instead of summing probabilities on all 21 cases where there's more primes than even numbers, it's easier to sum the 7 cases for E q .
| subcases ( ∙ =both prime and even, ∙ =not prime nor even) | |
| 0 even , 0 prime | ∙ ∙ ∙ ∙ ∙ ∙ ( 6 6 ) 6 6 1 = 6 6 1 |
| 1 even , 1 prime | ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ( 1 , 5 6 ) ⋅ 6 1 1 ⋅ 6 5 1 = 6 6 6 ( 1 , 1 , 4 6 ) ⋅ 3 1 1 ⋅ 3 1 1 ⋅ 6 4 1 = 6 6 1 2 0 |
| 2 even , 2 primes | ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ( 2 , 4 6 ) ⋅ 6 2 1 ⋅ 6 4 1 = 6 6 1 5 ( 1 , 1 , 1 , 3 6 ) ⋅ 6 1 1 ⋅ 3 1 1 ⋅ 3 1 1 ⋅ 6 3 1 = 6 6 4 8 0 ( 2 , 2 , 2 6 ) ⋅ 3 2 1 ⋅ 3 2 1 ⋅ 6 2 1 = 6 6 1 ′ 4 4 0 |
| 3 even , 3 primes | ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ( 3 , 3 6 ) ⋅ 6 3 1 ⋅ 6 3 1 = 6 6 2 0 ( 2 , 1 , 1 , 2 6 ) ⋅ 6 2 1 ⋅ 3 1 1 ⋅ 3 1 1 ⋅ 6 2 1 = 6 6 7 2 0 ( 1 , 2 , 2 , 1 6 ) ⋅ 6 1 1 ⋅ 3 2 1 ⋅ 3 2 1 ⋅ 6 1 1 = 6 6 2 ′ 8 8 0 ( 3 , 3 6 ) ⋅ 3 3 1 ⋅ 3 3 1 = 6 6 1 ′ 2 8 0 |
| 4 even , 4 primes | ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ( 4 , 2 6 ) ⋅ 6 4 1 ⋅ 6 2 1 = 6 6 1 5 ( 3 , 1 , 1 , 1 6 ) ⋅ 6 3 1 ⋅ 3 1 1 ⋅ 3 1 1 ⋅ 6 1 1 = 6 6 4 8 0 ( 2 , 2 , 2 6 ) ⋅ 6 2 1 ⋅ 3 2 1 ⋅ 3 2 1 = 6 6 1 ′ 4 4 0 |
| 5 even , 5 primes | ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ( 5 , 1 6 ) ⋅ 6 5 1 ⋅ 6 1 1 = 6 6 6 ( 4 , 1 , 1 6 ) ⋅ 6 4 1 ⋅ 3 1 1 ⋅ 3 1 1 = 6 6 1 2 0 |
| 6 even , 6 primes | ∙ ∙ ∙ ∙ ∙ ∙ ( 6 6 ) ⋅ 6 6 1 = 6 6 1 |
Total is E q = 6 6 9 ′ 0 2 4 = 2 4 3 4 7 , so M p = 2 1 ( 1 − 2 4 3 4 7 ) = 2 4 3 9 8 = 0 . 4 0 3 2 9 2 . . . .
Notation: ( p 1 , p 2 , ⋅ , p k n ) = p 1 ! ⋅ p 2 ! ⋅ ⋯ ⋅ p k ! n is the multinomial coefficient , which counts the ways to display n = p 1 + p 2 + ⋯ + p k elements when there is k groups of p 1 , ⋯ , p k identical elements.
notice that 2 is both prime and even hence it won't support either.
let c = primes - even.
thus 3,5 increase c by 1.
4 and 6 decrease c by 1.
1 and 2 do not change c. all the three groups have equal elements and hence equal probability. consider the generating function .. ( x 1 +1 +x) 6 .
in its expansion the sum ofs coefficient of negative powers will give the no of cases where c is negative , the constant term the no of cases where c=0 and the sum of coefficients of positive terms the no of cases where c is positive.
required probability is the no of cases where c is positive divided by total no of cases.
solving the generating function can be made easier by noticing that coefficients of negative powers = coefficients of positive powers. solving we have.
probability=294/3^6=294/729.