If girlfriend roll two fair six-sided dice, what is the probability the the amount is $4$ or higher?

The answer is $frac3336$ or $frac1112$. Ns understand how to come at this answer. What ns don"t recognize is why the price isn"t $frac911$? when summing the results after rolling two fair 6 sided dice, there are $11$ equally possible outcomes: $2$ v $12$. Two of these outcomes are listed below four, definition $9$ are higher than or equal to 4 which is exactly how I landed on $frac911$. Have the right to someone help explain why the is wrong?

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It is wrong due to the fact that it is not $11$ equally feasible outcome.

There is exactly $1$ way to acquire the sum to be $2$. ($1+1=2$)

but over there is an ext than one way to acquire $3$. ($1+2=3, 2+1=3$)


$egingroup$ To add to this, there space 36 equally feasible outcomes. 6 top top one die and 6 top top the other, 6 x 6 = 36. 1 of those outcomes is a 2, 2 that those outcomes is a 3 for this reason (36 - 3)/36 outcomes 4 or higher. $endgroup$
As mentioned by others, the feasible sums that the dice don"t all have an same probability of mirroring up. To see this, you deserve to write it every out:

Die 1 | die 2 | an outcome 1 1 2 1 2 3 1 3 4 ... 6 5 11 6 6 12When friend look in ~ the resulting table, there are 36 combinations. 1 the those is a 2 (ie you have actually a 1 in 36 opportunity of gaining a 2 indigenous D1 + D2), 2 of those are "3" etc.

Now it"s straightforward to see exactly how to obtain the opportunity of getting a 4 or higher.


$frac911$ is wrong precisely due to the fact that it assumes the probabilities of acquiring each sum are equal. Right here they space not: there"s just one means to roll a sum of 2 (the snake eyes of gambling jargon) but six ways to roll a seven.


if the correct answer is $ frac 3336$ it method there room 36 feasible results.. Friend should try to write them down..It will be more clear. You just used equiprobability in a instance where there isn"t equiprobability..$P(1)=0 quad P(2) =1/36,quad P(3)=2/36 ,quad P(4)=3/36,$

$ P(5)=4/36 quad P(6)=5/36 quad P(7)=6/36 quad P(8)=5/36 quad $

$ P(9)=4/36 quad P(10)=3/36 quad P(11)=2/36 quad P(12)=1/36$

And the answer is just:$$P=1-P(1)-P(2)-P(3)=frac 3336$$

Your probabilities space wrong. Because that each worth of dice 1, there space 6 feasible values of die 2, the obtainable scores because that a roll of two dice are: (7 is the most most likely score)

2,3,4,5,6,7 3,4,5,6,7,8 4,5,6,7,8,9 5,6,7,8,9,10 6,7,8,9,10,11 7,8,9,10,11,12So, out of the 36 total possible outcomes, 33 of them have a worth of 4 or higher, or 33:36, or 11:12

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