Find the binomial distribution V(x)= npq of tossing two dice together.
what do u want?
what will n, p and q be? i no n=1 but im little confused with p and q
i was thinking it is 2/12 or 1/6 because 2 possibilities out of 12
yes
i thought it was coin for a second
what is V(x)?
why is it a variable of x when n*p*q have no x in them
that is the equation
say it is rolling one die. would p= 1/6 and q=5/6
this is the probabillty of getting only 1 of the number u want
so say you want the number '1' on only 1 dice then| then prob = 1/6*5/6
so V(x) im guessing x = the number of times you want it to appear
yes
V(0)=5/6* 5/6 V(1)=1/6*5/6 V(2)=1/6 * 1/6
oh and multiply V(1)* 2
because u can also have the case where 5/6 * 1/6
i think let me think about that one for a sec
ok b/c i was wondering why the variance is the same for rolling one die and rolling two dice.
ok yep multiply by 2
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
umm???
n=number of trials p=probability of success q= probability of failure sorry im confusing you
what is this rolling one die and rolling 2 dice? you are always rolling 2 dice
do you mean landing the number you want once and landing it twice
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
oh u are rolling one die twice you mean?
as opposed to rolling 2 dice at the same time?
2 dice same time
im talking about the problem where u had only 1 die
how many times did u roll the 1 die...
once
lol then how are u getting 5/36 prob
dont get confused this stuff is very simple
you are too worried about the formulas
just think about it, if you roll a dice whats the chance u get the number you want
1/6...
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
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