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Mathematics 17 Online
OpenStudy (anonymous):

Find the binomial distribution V(x)= npq of tossing two dice together.

OpenStudy (dan815):

what do u want?

OpenStudy (anonymous):

what will n, p and q be? i no n=1 but im little confused with p and q

OpenStudy (anonymous):

i was thinking it is 2/12 or 1/6 because 2 possibilities out of 12

OpenStudy (dan815):

yes

OpenStudy (dan815):

i thought it was coin for a second

OpenStudy (dan815):

what is V(x)?

OpenStudy (dan815):

why is it a variable of x when n*p*q have no x in them

OpenStudy (anonymous):

that is the equation

OpenStudy (anonymous):

say it is rolling one die. would p= 1/6 and q=5/6

OpenStudy (dan815):

this is the probabillty of getting only 1 of the number u want

OpenStudy (dan815):

so say you want the number '1' on only 1 dice then| then prob = 1/6*5/6

OpenStudy (dan815):

so V(x) im guessing x = the number of times you want it to appear

OpenStudy (anonymous):

yes

OpenStudy (dan815):

V(0)=5/6* 5/6 V(1)=1/6*5/6 V(2)=1/6 * 1/6

OpenStudy (dan815):

oh and multiply V(1)* 2

OpenStudy (dan815):

because u can also have the case where 5/6 * 1/6

OpenStudy (dan815):

i think let me think about that one for a sec

OpenStudy (anonymous):

ok b/c i was wondering why the variance is the same for rolling one die and rolling two dice.

OpenStudy (dan815):

ok yep multiply by 2

OpenStudy (anonymous):

rolling one die: n=1 p=1/6 q=5/6 npq=5/36 rolling 2 dice: n=1 p=2/12 or 1/6 q= 10/12 or 5/6 npq=5/36

OpenStudy (dan815):

umm???

OpenStudy (anonymous):

n=number of trials p=probability of success q= probability of failure sorry im confusing you

OpenStudy (dan815):

what is this rolling one die and rolling 2 dice? you are always rolling 2 dice

OpenStudy (dan815):

do you mean landing the number you want once and landing it twice

OpenStudy (anonymous):

i had a problem where im rolling one die and another problem rolling 2 dice. i got the same answer for both so i was just wondering what im doing wrong

OpenStudy (dan815):

oh u are rolling one die twice you mean?

OpenStudy (dan815):

as opposed to rolling 2 dice at the same time?

OpenStudy (anonymous):

2 dice same time

OpenStudy (dan815):

im talking about the problem where u had only 1 die

OpenStudy (dan815):

how many times did u roll the 1 die...

OpenStudy (anonymous):

once

OpenStudy (dan815):

lol then how are u getting 5/36 prob

OpenStudy (dan815):

dont get confused this stuff is very simple

OpenStudy (dan815):

you are too worried about the formulas

OpenStudy (dan815):

just think about it, if you roll a dice whats the chance u get the number you want

OpenStudy (dan815):

1/6...

OpenStudy (dan815):

if you roll it again and u want the same number then 1/6 * 1/6 because only out of 1/6th of the first dice numbers will be urs and then another 1/6th of the 2nd dice

OpenStudy (dan815):

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