Question

A fair die

is thrown

seven

times.

Find

the probability

that

the outcome contains

one ‘1’,

two ‘2’, three ‘3’

, but

no

‘4’?

Answer 1

Clearly the sample will be of size 6⁷.

Now, we will count the number of ways one can arrange one 1, two 2s, three 3s :

So, basically we have 6 symbols of which two are same and the another three are same and one is single. Here the counting goes by the following 6!/(1! 2! 3!) =A (say)

Now we have only two symbols 5 and 6 to choose from, as we can't choose 4 and 1, 2 and 3 are done.

so we can make that choosing in 2 ways and we can put those two in 7 places as we can obtain 5 or 6 at any point of throw.

So, putting 5 or 6 in between the throws can be done in 2*7=14 ways.

So, total number of counting in our favour will be 14*A (A is defined as above)

So the required probability will be 14*A / (6⁷).