Question 476612: Hi I have included the problem for some reason I cannot attach the picture but If you can help me I will send the pic to you. I have tried to work this out but I am still needing a little bit more help so any assistance is really appreciated!!!
Question. Assume that one number from 1 to 7 is equally likely to be selected at random. Each number corresponds to one of the seven figures shown. Determine the probability of selecting a circle, given that an even number is selected.
Answer- The probability of selecting a circle given that an even number is selected = 2/3
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your number are from 1 to 7.
those numbers are 1,2,3,4,5,6,7
the odd number are:
1,3,5,7
the even numbers are:
2,4,6
the probability of getting an odd number is 4/7.
the probability of getting an even number is 3/7.
you say the answer is 2/3.
since you are given that the number is even, then the possible number are 2,4,6.
if 2 out of 3 of them contain a circle, then the probability would be 2/3.
i would guess that 2 out of the 3 even numbers contain a circle.
you can send me the picture, but if my assumption is correct, you don't need to.
this would fall under the heading of p(a given b) is equal to p(a intersect b) divided by p(b).
that would be shown as:
p(a|b) = p(a^b)/p(b)
^ means intersect.
| means given.
here's how you would calculate the probability using this formula.
p(a) is the probability of getting a circle.
p(b) is the probability of getting an even number.
the formula would read:
probability of getting a circle given that the number is an even number.
suppose your data was as follows:
1 = c
2 = c
3 = c
4 = c
5 = o
6 = o
7 = o
c = circle
o = other.
you have 4 circles in the data set.
2 of them are in odd numbers
2 of them are in even numbers.
p(a) is therefore 4/7 (4 circles out of a total of 7 numbers).
p(b) is 3/7 (3 even numbers out of a total of 7 numbers).
p(a^b) would be the intersection of even numbers with circles which would be 2/7 (there are 2 even numbers that contain circles out of a total of 7 numbers).
the formula is:
p(a|b) = p(a^b)/p(b)
that comes out to be:
p(a|b) = (2/7) / (3/7)
because (a/b)/(b/c) is equal to (a/b)*(c/b), that is equivalent to:
p(a|b) = (2/7) * (7/3)
the 7 cancels out and you are left with:
p(a|b) = 2/3
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