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[docveda]

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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!!!

 

[bluesdeluxe]

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?

 

[checkamundo]

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)

 

[checkamundo]

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

 

[docveda]

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!!!

 

[checkamundo]

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

 

[docveda]

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

cool