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Mathematics 12 Online
OpenStudy (anonymous):

You roll two fair dice, one green and one red. (b) Find P(5 on green die and 3 on red die). (c) Find P(3 on green die and 5 on red die). (d) Find P((5 on green die and 3 on red die) or (3 on green die and 5 on red die)). please help and explain with this... I'm stuck..

Nnesha (nnesha):

i'm not good at probability question but ehhh let me try it one dice = 6 sides you can get 5 only one time right and same with 3 = there is only possibility to get 3 on each dice |dw:1446522296655:dw|

OpenStudy (anonymous):

i appreciate that you're still trying to help (:

Nnesha (nnesha):

P(5G and 3R) \[\rm \color{green}{\frac{ 1 }{ 6}}\] possibility to get 5 \[\rm \color{red}{\frac{ 1 }{ 6 }}\] possibility to get 3 on one die

OpenStudy (anonymous):

yea.. so 2/24 ?

Nnesha (nnesha):

and since both dice are independent so multiply both fraction

Nnesha (nnesha):

now we should multiply them not add.

OpenStudy (anonymous):

oh so... 2/36?

Nnesha (nnesha):

\[\rm (\color{green}{\frac{ 1 }{ 6}})*(\color{red}{\frac{1}{6}})\] multiply numerator *numerator and denominator * denominator

OpenStudy (anonymous):

1/36..?

Nnesha (nnesha):

looks good

OpenStudy (anonymous):

think i got it.. thanks !

OpenStudy (anonymous):

mmmm what about the last one..?

Nnesha (nnesha):

for last one would do same thing first set up the fraction for `P((5 on green die and 3 on red die)` and for `(3 on green die and 5 on red die))`. and then add both result

Nnesha (nnesha):

the*

OpenStudy (anonymous):

so its 1/36 +1/36 ?

Nnesha (nnesha):

right

OpenStudy (anonymous):

thin i got it ... thank so much !

Nnesha (nnesha):

np :=))

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