Question

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

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

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

Answer 3

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

Answer 4

you need a calculator to continue

Answer 5

i get .172 rounded.

Answer 6

"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.

Answer 7

i will write them if you like

Answer 8

please show me, that is where I get lost

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

yes it is thanks

Answer 13

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 \[.2^1\times .8^{13}\]

Answer 14

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
\[P(X=1)=14\times .2^1\times .8^{13}\]

Answer 15

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

Answer 16

hello armistre

Answer 17

howdy ;)

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

i will do the next one too if you like

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

the answer for 3

Answer 29

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

Answer 30

yes.

Answer 31

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

Answer 32

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

Answer 33

yes you have it

Answer 34

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

Answer 35

i can write a general formula for you if you like

Answer 36

please

Answer 37

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

Answer 38

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

Answer 39

ok

Answer 40

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

Answer 41

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

Answer 42

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?

Answer 43

yes

Answer 44

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
\[p^k\times (1-p)^{n-k}\]

Answer 45

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

Answer 46

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 \[\dbinom{n}{k}\]

Answer 47

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:...
\[P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}\]

Answer 48

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

Answer 49

so k was the different numbers we used for side effects

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

what if they don't equal 1?

Answer 54

now sure what you are asking.

Answer 55

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

Answer 56

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
\[\frac{32}{36}\]

Answer 57

#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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

oh i see. that makes sense

Answer 64

good!

Answer 65

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

Answer 66

thank you you are awesome!

Answer 67

welcome!