If you throw a dice six times ,what's the chance that you get a six on exactly one of the throws ?

We are asked to find the probability of rolling exactly 1 six when rolling a standard die 6 times.

This is a binomial experiment: there are a fixed number of trials (6), there are only two outcomes (a 6 or not a 6), the outcomes are independent and the probability remains the same for each trial.

The formula is `([n],[x])p^xq^(n-x) ` where `([n],[x]) ` is the number of combinations of n objects taken x at a time, p is the probability of a success, and q is the complement of p (q=1-p). Here n=6, p=1/6, q=5/6, and x=1:

`([6],[1])(1/6)^1(5/6)^5=(5/6)^5=3125/7776~~.402 `

Thus P(x=1) is 3125/7776 or about .402

We could also get this from first principles:

There are 6 rolls. If only 1 is a 6, then there is 1 six with probability 1/6, and 5 other numbers each with probability 5/6. The 6 can occur in any of the six places so we have `6*(1/6)^1*(5/6)^5 ` as above.