With one fair die, find the probability of rolling two fours in five attempts.

I think this is a permuations problem. I just dont know how to solve it.

Thanks!!!

With one fair die, find the probability of rolling two fours in five attempts.

I think this is a permuations problem. I just dont know how to solve it.

Thanks!!!


one fair die being one normal 6 sided dice (die singular)?

seems like 5x6=30 and 2 fours / 30 possible = 1/15 change

correct?

With one fair die, find the probability of rolling two fours in five attempts.

I think this is a permuations problem. I just dont know how to solve it.

Thanks!!!


You have to use this eqn:

N!/M!(N-M)! x P^M x (1-P)^(N-M) where M= # of successes in N trials
and P= probability of occurance

So for your problem, the probability of throwing any number = P= 1/6

5!/2!(5-2)! x (1/6)^2 x (1-1/6)^(5-2)

that gives you an answer of 625/3888, i'm not sure if that's one of the answers you have. let me know if that's wrong

You have to use this eqn:

N!/M!(N-M)! x P^M x (1-P)^(N-M) where M= # of successes in N trials
and P= probability of occurance

So for your problem, the probability of throwing any number = P= 1/6

5!/2!(5-2)! x (1/6)^2 x (1-1/6)^(5-2)

yea you are right!! thanks!!!

yea you are right!! thanks!!!


By the way, the eqn is basically multiplying the combinations formula by the probability of successes, multiplied by the probability of failure

By the way, the eqn is basically multiplying the combinations formula by the probability of successes, multiplied by the probability of failure

cool