suppose a balanced die is tossed 10 times what is the probability that all the faces 6 faces appear at least once

Question

suppose a balanced die is tossed 10 times what is the probability that all the faces 6 faces appear at least once

Answer 1

Absolutely wrong answer.

Try this approach for 6 tosses and you will get negative probability.

Answer 2

WRONG ANSWER: (Correct one is below)

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The best way to calculate this probability is to calculate the probability when it is not happening and subtracting from 1.

The probability of not happening:

prob(x do not appear for all 10 tosses) * 6

where x is any number on the surfaces of the die.

Our condition is met except the probability above. So,

The probability of happening:

1 - [prob(x does not appear for all 10 tosses) * 6]

And,

prob(x does not appear for one toss) = 5/6

prob(x does not appear for ten tosses) = (5/6) ^ 10

The probability of happening of the condition:

1 - {[(5/6) ^ 10] * 6} ~= 0.031

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UPDATE: Above solution is wrong. It misses the calculation of no appearence of more than 1 numbers for all 10 tosses.

CORRECT ANSWER:

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So new approach is:

we need each number for at least one toss,

prob(x for one toss) = 1/6,

for ten tosses if we have 1, 2, 3, 4, 5, and 6 for in first six tosses the rest of tosses does not matter for us, so the probability of these unique case is,

[(1 / 6) ^ 6] * 1 ^ 4

the probability of all cases for our condition is all combinations of these tosses,

C(10,6) * [(1 / 6) ^ 6] ~= 0.0045

{What about this Valentin?}