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Mathematics 44 Online
OpenStudy (anonymous):

A number is chosen from a list 2,4,6 and 8. Another is selected from from the list 6,7 and 8. Find the probability that they are the same.

OpenStudy (shubhamsrg):

req probability = (2/4) * (2/3)

OpenStudy (shubhamsrg):

as 2 nos are a part if favorable outcomes from both lists..

OpenStudy (anonymous):

I don't really get what you just said lol

OpenStudy (anonymous):

good because i am pretty sure it is not right

OpenStudy (anonymous):

Good.

OpenStudy (shubhamsrg):

from 1st list, there are 4 nos : 2,4,6,8 in 2nd : 6,7,8 2 nos i.e. 6 and 8 are common to both we select a no. each from the list.. probability of selecting 6 or 8 from 1st list = 2/4 from 2nd list = 2/3 we want both to happen simultaneously, so req probab = 2/4 * 2/3

OpenStudy (shubhamsrg):

ohh..please correct me.

OpenStudy (anonymous):

picture of problem

OpenStudy (anonymous):

there are two outcomes favorable both are 6 or both are 8 the probability both are 6 is \[\frac{1}{4}\times \frac{1}{3}=\frac{1}{12}\] the probability both are 8 is the same,\(\frac{1}{12}\) you want one of the other, and since these events (both 6 or both 8) are mutually exclusive, you can add the probabilities together

OpenStudy (anonymous):

so you just add them together?

OpenStudy (anonymous):

yes just add

OpenStudy (anonymous):

because the events are "disjoint" simply meaning you cannot get two 6 and at the same time two 8 in such a case an "or" statement i.e. one or the other you compute by adding

OpenStudy (shubhamsrg):

ohh yes yes,,i see what i did..i even included (6,8) and (8,6) in my ans sorry and thank you @satellite73 hmmm

OpenStudy (anonymous):

you can think of it this way there are 12 outcomes all together which just 2 of them are favorable so the answer is 2/12 = 1/6

OpenStudy (shubhamsrg):

yep..get it now @Kasra sorry again.. :|

OpenStudy (anonymous):

thanks!

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