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.
req probability = (2/4) * (2/3)
as 2 nos are a part if favorable outcomes from both lists..
I don't really get what you just said lol
good because i am pretty sure it is not right
Good.
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
ohh..please correct me.
picture of problem
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
so you just add them together?
yes just add
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
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
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
yep..get it now @Kasra sorry again.. :|
thanks!
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