# Homework Help: Probability of a dice

1. Jan 26, 2010

### Dell

i throw a dice 5 times,
1)what is the probability that i get the number 6 at least twice?

2)what is the probability that i get the same number at least 4 times?

i know that since i am throwing the dice 5 times i have 6^5=7776 options
now all i need to find out is how many of those options contain 2 or more sixes, adn the same number 4 or more times
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1) the number of options containing 6 twice

i think it would be easier to find the amount that contain 0 or 1 six and subtract.
since 0 or 1 of the rolls must be 6 and the rest must not be 6
1*5*5*5*5 + 5*5*5*5*5 = 3750

giving me the probability 3750/7776=0.482 that i will roll less than 2 sixes
and the probability of 0.518 that i will roll 2 or more

but the answer is meant to be 0.196?? where am i going wrong?
had i used 5*5*5*5*5 + 5*5*5*5*5 =6250 instead of 3750 i would have gotten the right answer, but i doint understand why that is correct.

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2) to get the same number 4 or more times i need to get the same number either 4 times or 5 times
the first number can be any number, the next 3 must be the same as the first and the last must either be the same as the first or different to the first
6*1*1*1*1 + 6*1*1*1*5=36

36/7776=0.0046

but again the correct answer is meant to be 0.02 and i cannot see how,

2. Jan 26, 2010

### vela

Staff Emeritus
You didn't take into account the number of ways to roll a 6 only once. You have five possibilities: 6XXXX, X6XXX, XX6XX, XXX6X, XXXX6.

You made the same error on the second problem.

3. Jan 26, 2010

### Dell

okay i think i got it

5*5*5*5*5+(1*5*5*5*5)*5, and that really gives me 0.1962

but for the second one

(6*1*1*1*1)+(6*1*1*1*5) but do i multiply this by 5 also?? are there not more than 5 possibilities for the second term?

4. Jan 26, 2010

### Dell

also in another question which i thought i had right i was asked the possibility to roll the same number 3 times exactly out of 6

i said

6*1*1*5*5 since the first roll i can have any number, 2 others must be the same and 2 others must be different, but here too i didnt take into account the ways i could roll these numbers, i could have

6 1 1 5 5
1 1 5 5 6
1 5 5 6 1
5 5 6 1 1
5 6 1 1 5
1 6 1 5 5
6 1 5 5 1
6 1 ......................

it seams like it never going to end, what os the method to solve this?