We want to use the following probabilities to answer the question.
It may be helpful to sketch a Venn diagram, which we will do.
We're given the probability of A equals zero point five, six.
The probability of B equals zero point three, four,
and the probability of A and B equals zero point one zero.
We want to find the probability of not B
given A has already occurred. This is conditional probability which is defined
below.
Applying our definition to find this conditional probability,
we need to find the probability of A
and not B,
divided by
the probability of A. Now let's go ahead and complete our Venn diagram
using the given probabilities. We'll start with the intersection of A and B.
The probability of A and B equals zero point one,
which would be the probability of being in this region here.
Next because the probability of A
equals zero point five, six, it does include this probability because notice how this
region is in A.
So the probability of being in this region, which is only A
would be zero point five, six minus zero point one,
which is zero point four, six And the probability of B equals zero point three, four.
And again this is included in the zero point three, four
so the probability of only B would be zero point three, four
minus zero point one, or zero point two, four.
Now let's also find the probability of being in this outer region
which would be the probability of being in the universal set,
but not in A, and not in B. This would be equal to
one minus, the sum of these three probabilities,
which would be zero point two. Again this is the probability of being
not in A and not in B. Of course we can check this on the calculator.
One minus the quantity point four six,
plus point one, plus
point two, four does give us
zero point two. And now before we find these two probabilities,
let's highlight the region A, and not B, as well as A.
Well let's start with the easy one, the region A is this region here.
Because of the conditional probability,
that says not B, given A, we can already assume we're in this region here.
And the tricky part here is what region is A
and not B? So it has to be in A
and not in B, which would be the region that's only A, or this region here.
So notice how to find this conditional probability,
we take the probability of A and not B, which is the probability of only A, which is
zero point four, six.
And we divide by the probability of A
which we know was given at zero point five, six. It is also the sum of these two probabilities.
So again what happened here, to find this conditional probability,
we took the probability of the desired region
highlighted here in orange, and divided by the probability of the total region
given that we already knew we were in A, which was the probability of A.
Let's first clear the decimals from our fraction,
so we can multiply by one hundred over one hundred.
Which would give us
forty-six, fifty-six, but now we have a common factor of two,
so we can divide the numerator and denominator by two.
So the simplified probability as a fraction would be twenty-three,
twenty-eighths. Let's also write this as a decimal
and percentage. To convert to decimal we divide,
so twenty-three divided by twenty-eight
to four decimal places would be approximately zero point eight,
two, one four.
To convert to a percentage, we multiply by one hundred, add a percent sign,
or move the decimal point to the right two places and a percent sign,
giving us eighty-two point one, four percent.
Notice how the exact probability would be the fraction,
because these two around it, these would be approximations.
So for our homework, we should enter the exact probability as twenty-three, twenty-eighths.
But also recognize that often probability is written as a decimal,
or percentage.
I hope you found this helpful.