In testing a new drug, researchers found that 20% of all patients using it will have a mild side effect. A random sample of 14 patients using the drug is selected. Find the probability that:
(A) exactly four will have this mild side effect
(B) at least five will have this mild side effect.
Help, I am failing this class

- dboyette

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- anonymous

exactly 4:
4 will have it, ten will not so your probability is
\[P(x=4)=\dbinom{14}{4}.2^4\times .8^{10}\]

- anonymous

where \[\dbinom{12}{4}\] is the number of ways you can choose 4 people from a set of 14.
\[\dbinom{14}{4}=\frac{14\times 13\times 12\times 11}{4\times 3\times 2}=7\times 13\times 11=1001\]

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## More answers

- anonymous

you need a calculator to continue

- anonymous

i get .172 rounded.

- anonymous

"at least 5" is a real pain because you either have to compute the probabilities for X = 0, 1, 2, 3, and 4 and then subtract them from 1.

- anonymous

i will write them if you like

- dboyette

please show me, that is where I get lost

- anonymous

ok. we put X = number of cases of side effects. you are given that the probabiltiy any one person has it is 20%=.2 so the probability any one person does not have it is 80% = .8

- anonymous

easy to compute the probability no one has it, because it is just \[.8^{14}\]

- anonymous

in other words you get 14 in a row without the side effect. is that much clear?

- dboyette

yes it is thanks

- anonymous

ok now let us compute the probability that one person has the side effect. one person has it, that means 13 do not. so it is going to look like \[.2^1\times .8^{13}\]

- anonymous

but there are 14 different ways for this to happen. either the first person has the side effect, or the second , or the third , or so on.
so the probabily that on person has it is
\[P(X=1)=14\times .2^1\times .8^{13}\]

- dboyette

so P(X=2)= 14*.2^2*.8^12?

- anonymous

hello armistre

- amistre64

howdy ;)

- anonymous

bdoyette you are close. one part will be \[.2^2\times .8^{12}\]

- anonymous

but this time you don't multiply by 14, you multiply by the number of ways you can choose 2 people out of 14

- anonymous

this is usually written as \[\dbinom{14}{2}\] and is computed as follows:
\[\dbinom{14}{2}=\frac{14\times 13}{2}=7\times 13=91\]

- anonymous

so
\[P(X=2)=91\times .2^2\times .8^{12}\]

- anonymous

i will do the next one too if you like

- dboyette

so the next one would be 14/3=14*13*12/3?

- anonymous

\[\dbinom{14}{3}=\frac{14\times 13\times 12}{3\times 2}\]

- anonymous

think as follows. I pick three from a set of 14. i have 14 choices for the first, 13 for the second and 12 for the third. this explains the numerator. but since i don't care what order i pick them in i have counted too many ways. the numerator counts (ABC) different than (ACB) different than (BAC) etc. so i have to divide by how many ways i can permute these three people which is 3*2=6

- anonymous

another example would be \[\dbinom{14}{5}=\frac{14\times 13\times 12\times 11\times 10\times 9}{5\times 4\times 3\times 2}\]

- dboyette

so that would be 364* 0.2^3*0.8^11?

- dboyette

the answer for 3

- anonymous

yes i think i didn't compute it so let me check

- anonymous

yes.

- dboyette

the # 5 would be 18018*0.2^5*0.8^9?

- anonymous

\[\dbinom{14}{3}=\frac{14\times 13\times 12}{3\times 2}=14\times 13\times 2=364\]

- anonymous

yes you have it

- dboyette

omg i think i am finally starting to get this, thank you so much

- anonymous

i can write a general formula for you if you like

- dboyette

please

- anonymous

ok you in this example you have a probability that is 20%=.2
in the general case you would have a probability that is a variable, call it p

- anonymous

we knew that if the probability of having a side effect was .2 then the probability of not having it was .8 because they have to add to 1

- dboyette

ok

- anonymous

so in the general case we would have the probability of having it is p, and the probability of not having it is 1-p

- anonymous

in this case we had 14 people, in the general case we will have n people (tosses of a coin, whatever)

- anonymous

so if i want to know the probability that 5 people have the side effect, that means that 9 don't. in the general case if i want to know the probability that k people have it that means that n-k will not. so far so good?

- dboyette

yes

- anonymous

ok now we computed the probability that 5 people had it by computing \[.2^{5} \times .8^9\] and then multiplying by the number of ways to select 5 people from 9. in the general case we will have k people with the side effect and n - k without, so you will have to compute
\[p^k\times (1-p)^{n-k}\]

- anonymous

that is like our \[.2^5\times .8^9\]

- anonymous

but then you will have to multiply by the number of ways you can choose those 5 out of 14, or in this case the number of ways you can choose k out of n. that is written as \[\dbinom{n}{k}\]

- anonymous

so the "final answer" is, if you have n independent trials with probability of success = p and therefore the probability of failure = 1-p at each trial (these are called bernoulli trials) the the probability you get exactly k success is:...
\[P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}\]

- anonymous

we computed this with n = 14, p = .2 and 1-p=.8 for various different values of k

- dboyette

so k was the different numbers we used for side effects

- anonymous

yes, the number of people who could have it. question one was for k = 4

- anonymous

to do question 2 you have to compute the probability
\[P(X=0)\] which we did, the probability
\[P(X=1)\] which we also did, the probability
\[P(X=2)\]
\[P(X=3)\]
\[P(X=4)\] a real pain. then we add these numbers up and subtract them from 1

- anonymous

because "5 or more" means "not none, not one, not two, not three and not four"

- dboyette

what if they don't equal 1?

- anonymous

now sure what you are asking.

- anonymous

the total is 1 for all the possibilities. it has to be, because the probability of the whole sample space is always one.

- anonymous

just like if i know that the probability i roll a 10 on two dice is \[\frac{4}{36}\] then i know the probability i don't is
\[\frac{32}{36}\]

- dboyette

#2 91*.2^2*.8^12 ==0.25
#3 364*0.2^3*0.8^11 = 0.25
#4 1001*0.2^4*0.8^10 = 0.17
5# 18018*0.2^5*0.8^9=0.77
these don't add up to 1

- anonymous

ok let me go slow. the total is one, meaning that if you compute all the probabilities
for x = 0, x =1, x = 2, x= 3, x = 4, x = 5, x = 6, x = 7, x = 8, x = 9, x = 10, x = 11, x = 12, x = 13, x = 14 and add them up you will get 1

- anonymous

you only want at least 5, meaning x = 5, x = 6, x = 7, x = 8, x = 9, x = 10 , x = 11, x = 12, x = 13, x = 14

- anonymous

you can compute all these if you want, and add them and that will be your answer.

- anonymous

but it is easier to compute the probabilities for x = 0, x = 1, x = 2, x = 3, and x = 4. add them up and then subtract this number from 1

- anonymous

this way you only have to compute 5 probabilities instead of 10

- dboyette

oh i see. that makes sense

- anonymous

good!

- anonymous

btw you do not need to compute \[P(X=5)\] because it says "at least 5" so you compute just the lower ones and subtract the result from 1

- dboyette

thank you you are awesome!

- anonymous

welcome!

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