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anonymous
 5 years ago
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
anonymous
 5 years ago
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

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! 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!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0exactly 4: 4 will have it, ten will not so your probability is \[P(x=4)=\dbinom{14}{4}.2^4\times .8^{10}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where \[\dbinom{12}{4}\] is the number of ways you can choose 4 people from a set of 14. \[\dbinom{14}{4}=\frac{14\times 13\times 12\times 11}{4\times 3\times 2}=7\times 13\times 11=1001\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need a calculator to continue

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"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
 5 years ago
Best ResponseYou've already chosen the best response.0i will write them if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please show me, that is where I get lost

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. 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
 5 years ago
Best ResponseYou've already chosen the best response.0easy to compute the probability no one has it, because it is just \[.8^{14}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in other words you get 14 in a row without the side effect. is that much clear?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0but 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 \[P(X=1)=14\times .2^1\times .8^{13}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so P(X=2)= 14*.2^2*.8^12?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bdoyette you are close. one part will be \[.2^2\times .8^{12}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but this time you don't multiply by 14, you multiply by the number of ways you can choose 2 people out of 14

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this 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
 5 years ago
Best ResponseYou've already chosen the best response.0so \[P(X=2)=91\times .2^2\times .8^{12}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i will do the next one too if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the next one would be 14/3=14*13*12/3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\dbinom{14}{3}=\frac{14\times 13\times 12}{3\times 2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0think 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
 5 years ago
Best ResponseYou've already chosen the best response.0another example would be \[\dbinom{14}{5}=\frac{14\times 13\times 12\times 11\times 10\times 9}{5\times 4\times 3\times 2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so that would be 364* 0.2^3*0.8^11?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i think i didn't compute it so let me check

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the # 5 would be 18018*0.2^5*0.8^9?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\dbinom{14}{3}=\frac{14\times 13\times 12}{3\times 2}=14\times 13\times 2=364\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0omg i think i am finally starting to get this, thank you so much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can write a general formula for you if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0we 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so in the general case we would have the probability of having it is p, and the probability of not having it is 1p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in this case we had 14 people, in the general case we will have n people (tosses of a coin, whatever)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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 nk will not. so far so good?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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 \[p^k\times (1p)^{nk}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is like our \[.2^5\times .8^9\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0so the "final answer" is, if you have n independent trials with probability of success = p and therefore the probability of failure = 1p at each trial (these are called bernoulli trials) the the probability you get exactly k success is:... \[P(X=k)=\dbinom{n}{k}p^k(1p)^{nk}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we computed this with n = 14, p = .2 and 1p=.8 for various different values of k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so k was the different numbers we used for side effects

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, the number of people who could have it. question one was for k = 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to 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
 5 years ago
Best ResponseYou've already chosen the best response.0because "5 or more" means "not none, not one, not two, not three and not four"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what if they don't equal 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now sure what you are asking.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the total is 1 for all the possibilities. it has to be, because the probability of the whole sample space is always one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just 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 \[\frac{32}{36}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0#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
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0you can compute all these if you want, and add them and that will be your answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0this way you only have to compute 5 probabilities instead of 10

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see. that makes sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0btw 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you you are awesome!
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