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Mathematics 20 Online
OpenStudy (anonymous):

You roll 2 dice. What is the probability that the sum of the dice is less than 7 and at least 1 die shows a 3? (A 6 6 table of dice outcomes will help you to answer this question.)

OpenStudy (anonymous):

this is a hard one

OpenStudy (anonymous):

no

OpenStudy (anonymous):

for me it is.

OpenStudy (anonymous):

tell me how to do it.

OpenStudy (anonymous):

dont just give me the answers...

OpenStudy (anonymous):

its 1/12

OpenStudy (anonymous):

the probability you role a 3 is 1/6th

OpenStudy (anonymous):

now that you have 3 in order for the sum to be less then 7, the second dice must role less then 4

OpenStudy (anonymous):

the probability you role a dice with less then 4 is 1/2

OpenStudy (anonymous):

there mutually independent events, so the overall probability is 1/12

OpenStudy (anonymous):

thats not an answer choice...

OpenStudy (anonymous):

what are the answer choices

OpenStudy (anonymous):

7/12 1/3 5/36 7/36

OpenStudy (anonymous):

order counts?

OpenStudy (anonymous):

does the 3 have to appear in the first role?

OpenStudy (anonymous):

i believe so

OpenStudy (fifciol):

the probability is 5/36

OpenStudy (anonymous):

ye im to lazy to figure this out good luck

OpenStudy (anonymous):

proof pls

OpenStudy (anonymous):

5/36 because the first roll mus be a 3

OpenStudy (fifciol):

you have only 5 possibilities: first dice second dice 3 1 3 2 1 3 2 3 and 3 3 out of 6*6 possibilities of rolling two dices

OpenStudy (anonymous):

yeah i think hes right

OpenStudy (anonymous):

oh I see what you did there

OpenStudy (anonymous):

count the number of integers k on a die so that 3+k<7

OpenStudy (anonymous):

and multiply it by two for double pairs and subtract 1 so you don't over count the 3,3

OpenStudy (anonymous):

and divided by 36

OpenStudy (fifciol):

it could be done in that way but i prefer to simply write down the possibilities and count it when youre dealing with dice

OpenStudy (anonymous):

Fifciol you study mathematics

OpenStudy (anonymous):

i think i got it wrong

OpenStudy (anonymous):

thanks a lot..

OpenStudy (anonymous):

no problem,

OpenStudy (anonymous):

study any analysis fifciol?

OpenStudy (fifciol):

mathematics is just a tool to solve problems but dont you want to make life easier? :)

OpenStudy (fifciol):

of course

OpenStudy (anonymous):

thats narrow minded lol, do you know about special functions?

OpenStudy (anonymous):

like zeta functions/l functions etc

OpenStudy (fifciol):

yes i do

OpenStudy (anonymous):

do you know about euler products, those are nice

OpenStudy (anonymous):

like $$\prod_{p}(1+f(p)+f(p^2)+f(p^3)....)=\sum_{n=1}^\infty f(n)$$

OpenStudy (anonymous):

or is that to number theorish for u

OpenStudy (fifciol):

i dealt with gamma function and i remember i solved those integrals for some simple numbers

OpenStudy (anonymous):

nvm, good luck with life or whatever

OpenStudy (fifciol):

do you believe that everything is written in numbers?

OpenStudy (anonymous):

that is kinda vague, what do you mean by that

OpenStudy (anonymous):

heres a nice telescoping sum, $$\frac{1}{\ln(q)}=\sum_{n=-\infty}^\infty\frac{2^n}{q^{2^n}+1}$$

OpenStudy (anonymous):

i am a bit random

OpenStudy (fifciol):

you gotta be more pragmatic

OpenStudy (anonymous):

pragmatic?

OpenStudy (anonymous):

its 2:36 am in california, where are you?

OpenStudy (fifciol):

we're talking about problem with dice you're giving the wrong answer im trying to make you see it and then you show me a telescopic sum and tell that jack is not your name doont you think it kinda strange?

OpenStudy (fifciol):

i am euler and its 99:99 in math heaven :)

OpenStudy (fifciol):

it is one :) i got You :)

OpenStudy (fifciol):

yes

OpenStudy (anonymous):

live in a dorm?

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