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Mathematics 18 Online
OpenStudy (anonymous):

HELP ASAP WILL GIVE MEDAL AND BE FAN If Greg rolls four dice, what is the probability that he rolls more 1's than 6's?

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

u know what will be the individual probability??

OpenStudy (anonymous):

What is that?

OpenStudy (anonymous):

like if u roll one dice

OpenStudy (anonymous):

what is the probablity if u roll one dice for the given constraints?

OpenStudy (anonymous):

If you roll one dice, then you can't have more. There's only one number to base it on.

OpenStudy (anonymous):

first find prob of rolling 1

OpenStudy (anonymous):

then prob of rolling 6

OpenStudy (anonymous):

1/6

OpenStudy (anonymous):

I am little rusty in probability...hold on

OpenStudy (anonymous):

1 dice - probability of any number is 1/6 4 dices - probability of any number is 4/6 so the probability is the same for any number

OpenStudy (anonymous):

Does this mean that the probability is 4/6?

OpenStudy (anonymous):

@Fallontine it says how many more 1s than 6s??

OpenStudy (anonymous):

how do u interpret that??

OpenStudy (anonymous):

ammm... wait...

OpenStudy (anonymous):

ok..:)

OpenStudy (anonymous):

this is a permutation

OpenStudy (anonymous):

oh......so how did u proceed??

OpenStudy (anonymous):

prob of finding 1=1/6 of finding 6=5/6

OpenStudy (anonymous):

for one dice

OpenStudy (anonymous):

noooo... it's 1/6 for 1, 1/6 for 2 etc... and 1/6 for 6 coz you have 6 numbers and you have to choose 1 and the probability is the same for all of them

OpenStudy (anonymous):

I know, what permutation is this??

OpenStudy (anonymous):

Wait, can't we do casework?

OpenStudy (anonymous):

We can consider the various cases where there are more ones than sixes - one dice with a 1 and no dice with 6s - two dice with 1s and no dice with 6s - two dice with 1s and one dice with 6 - three dice with 1s and no dice with 6s - three dice with 1s and one dice with 6 - four dice with 1s

OpenStudy (anonymous):

no no after u have chosen 1/6 for 1 6 can be anywhere in the rest 5 numbers hence 5/6

OpenStudy (anonymous):

Is that right??

OpenStudy (anonymous):

@Fallontine

OpenStudy (anonymous):

Hello? :)

OpenStudy (turingtest):

it seems you just need to find how many ways the outcome can have more 1s than 6s and divide that by the total number of possible outcomes

OpenStudy (anonymous):

Oh yeah! Just a sec while I figure it out.

OpenStudy (anonymous):

1111 1116 1161.....like that

OpenStudy (anonymous):

@TuringTest Thank u .I got it

OpenStudy (anonymous):

How many possible outcomes are there? @TuringTest

OpenStudy (turingtest):

1116, 1161, 1611, 6111, 1111 five outcomes with more 1s than 6s

OpenStudy (anonymous):

I mean the number in the denominator.

OpenStudy (turingtest):

well how many total ways can one throw the dice? there are 6 possible outcomes for the first die 6 outcomes for the second die, etc

OpenStudy (anonymous):

Oh, so 6^4

OpenStudy (turingtest):

when you flip a coin, the number of different outcomes you can get is 2^n where n is the number of flips yep :)

OpenStudy (anonymous):

you also have to count with the other numbers (2 to 5) there are many many possibilities with no 1 or 6 at all

OpenStudy (turingtest):

indeed, I forgot to count 11x6, 1xxx, and so on my mistake

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