Probability Maths problem help.?

Obviously, the chance should be significantly low, so the first answerer is wrong.

First, we must know that by factorizing 10^99: we have 2^99 x 5^99

And factorizing 10^94: we have 2^94 x 5^94

Probability that a divisor of 10^99 will be a multiple of 10^94 is must have at least 94 multiplicities of 2, and at least 94 multiplicities of 5.

Let C(AAA ZZZ's) denote the case of AAA (times) of multiplicities of ZZZ, for example: P(5 6's) is the case of 5 multiplicities of 6's.

And C(BBB CCC, DDD EEE) denote the number of BBB (times) of multiplicities of CCC, and the number of DDD (times) of multiplicities of EEE. So similarly, C(3 4's, 5 6's) is the caseof 3 multiplicities of 4's and 5 multiplicities of 6's.

So the cases is:

C(94 2's, 94 2's) + C(94 2's, 95 2's) + C(94 2's, 96 2's) + ... + (94 2's, 99 2's)

+ C(95 2's, 94 2's) + C(95 2's, 95 2's) + C(95 2's, 96 2's) + ...+ (95 2's, 99 2's)

+ C(96 2's, 94 2's) + C(96 2's, 95 2's) + C(96 2's, 96 2's) + ...+ (96 2's, 99 2's)

+ C(97 2's, 94 2's) + C(97 2's, 95 2's) + C(97 2's, 96 2's) + ...+ (96 2's, 99 2's)

+ C(98 2's, 94 2's) + C(98 2's, 95 2's) + C(98 2's, 96 2's) + ...+ (96 2's, 99 2's

+ C(99 2's, 94 2's) + C(99 2's, 95 2's) + C(99 2's, 96 2's) + ...+ (96 2's, 99 2's)

[ I think you can see the pattern here]

P(a divisor of 10^99 will be a multiply of 10^94)

= nCr(99,94) * nCr(99, 94)

+ 2 nCr(99, 94) * nCr(99, 95)

+ 2 nCr(99, 94) * nCr(99, 96)

+ 2 nCr(99, 94) * nCr(99, 97)

+ 2 nCr(99, 94) * nCr(99, 98)

+ 2 nCr(99, 94) * nCr(99, 99)

+ nCr(99,95) * nCr(99, 95)

+ 2 nCr(99,95) * nCr(99, 96)

+ 2 nCr(99,95) * nCr(99, 97)

+ 2 nCr(99,95) * nCr(99, 98)

+ 2 nCr(99,95) * nCr(99, 99)

+ nCr(99,96) * nCr(99, 96)

+ 2 nCr(99,96) * nCr(99, 97)

+ 2 nCr(99,96) * nCr(99, 98)

+ 2 nCr(99,96) * nCr(99, 99)

+ nCr(99,97) * nCr(99, 97)

+ 2 nCr(99,97) * nCr(99, 98)

+ 2 nCr(99,97) * nCr(99, 99)

+ nCr(99,98) * nCr(99, 98)

+ 2nCr(99,98) * nCr(99, 99)

+ nCr(99,99) * nCr(99, 99)

All divided by (2^99 * 5^99)

= 3,557,874,930,289 / (625 x 10^93)

≈ 5.6925998884624 × 10^-84

[So extremely super duper unlikely for it will happen ]

Links to the calculations, in parts:

http://www.wolframalpha.com/input/?i=nCr%2899%2C94...

http://www.wolframalpha.com/input/?i=%2B+nCr%2899%...

http://www.wolframalpha.com/input/?i=%2B+nCr%2899%...

http://www.wolframalpha.com/input/?i=%2B+nCr%2899%...

http://www.wolframalpha.com/input/?i=%285677185030...

Break down the Divisor

10^99 = (2*5) ^99

So you have 99 2s and 99 5s.

Totaling 198 multiples.

10^94 = (2*5) ^ 94

94 2s and 94 5s = 188

The other 10^5 are high than 10^94 and therefore not multiples of 10^94

You get the probability 188/198 = 94/99.

Answer: 94/99