How to start math problem help please!

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sandi

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I am trying to figure out how to approach the two following Multiplecation Rules problems, any help would be wonderful!

1. A national study of patients who were overweight found 56% also have elevated blood pressures. If two overweight patients are selected, find the probability
that both have elevated blood pressure.


2. A flashlight has six batteries, two of which are defective. If two are selected at random without replacement, find the probability that the first battery tests good and the second one is defective.
 
2. A flashlight has six batteries, two of which are defective. If two are selected at random without replacement, find the probability that the first battery tests good and the second one is defective.
Click to expand...


The probability of choosing a good one on the first draw would be 4/6. Right?. There are 4 good ones and 2 bad ones.

Then, since there is no replacement, there are 5 to choose from and two of those are bad. So, the probability of choosing the bad one would be

2/5.
 
sandi said:
1. A national study of patients who were overweight found 56% also have elevated blood pressures. If two overweight patients are selected, find the probability
that both have elevated blood pressure.
Click to expand...

The probability the first has elevated blood pressure is 0.56,
same for the 2nd, (on average)

so the probability they both have elevated blood pressure is (0.56)0.56.
 
I am still confused on how you came up with 4/6 for the first answer, If you use the formula P(A and b)= P(A)* P(B\A) wouldn't the probability be 2/6 and 2/5 with the answer being 4/30 or 2/15?
 
Because we want the probability the first battery is good. There are 4 good ones and two bad ones in the 6.

So, the probability we choose a good one from the 6 is 4/6=2/3.

\(\displaystyle P(\text{first good AND second bad})=P(\text{first good})\cdot P(\text{second bad given first is good})\)

\(\displaystyle (\frac{4}{15})=(\frac{4}{6})(\frac{2}{5})\)
 
Ok that makes more sence to me, Thank you! Now would the same be for the following:?

In a scientific study, there are 8 hamsters, 5 of which have black fur. If 3 are selected at random without replacement, find the probability that all have black fur.

I am really struggling with the multiplication rules any pointers on how to read the problems would be much appreciated!!!
 
Yes, the idea is similar, sandi.

The chance the first one you pick having black fur is 5/8,
because 5 of the 8 you have to choose from have black fur.

The chance of the 2nd one having black fur is 4/7,
since 4 of the remaining 7 have black fur.

Then for the 3rd, the chance is 3/6.

Another way to think it through is...

There are 8(7)6 ways to pick 3 hamsters.
There are 5(4)3 ways to pick 3 hamsters with black fur.

Hence the probability of picking 3 hamsters with black fur is \(\displaystyle \frac{5(4)3}{8(7)6}\)
 
sandi said:
I am trying to figure out how to approach the two following Multiplecation Rules problems, any help would be wonderful!

1. A national study of patients who were overweight found 56% also have elevated blood pressures. If two overweight patients are selected, find the probability
that both have elevated blood pressure.


2. A flashlight has six batteries, two of which are defective. If two are selected at random without replacement, find the probability that the first battery tests good and the second one is defective.
Click to expand...

Thanks you for the post.
Hi guys, Im a newbie. Nice to join this forum.
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