dboyette
  • dboyette
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Owlfred
  • Owlfred
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anonymous
  • 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
  • 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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anonymous
  • anonymous
you need a calculator to continue
anonymous
  • anonymous
i get .172 rounded.
anonymous
  • 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
  • anonymous
i will write them if you like
dboyette
  • dboyette
please show me, that is where I get lost
anonymous
  • 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
  • anonymous
easy to compute the probability no one has it, because it is just \[.8^{14}\]
anonymous
  • anonymous
in other words you get 14 in a row without the side effect. is that much clear?
dboyette
  • dboyette
yes it is thanks
anonymous
  • 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
  • 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
  • dboyette
so P(X=2)= 14*.2^2*.8^12?
anonymous
  • anonymous
hello armistre
amistre64
  • amistre64
howdy ;)
anonymous
  • anonymous
bdoyette you are close. one part will be \[.2^2\times .8^{12}\]
anonymous
  • 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
  • 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
  • anonymous
so \[P(X=2)=91\times .2^2\times .8^{12}\]
anonymous
  • anonymous
i will do the next one too if you like
dboyette
  • dboyette
so the next one would be 14/3=14*13*12/3?
anonymous
  • anonymous
\[\dbinom{14}{3}=\frac{14\times 13\times 12}{3\times 2}\]
anonymous
  • 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
  • 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
  • dboyette
so that would be 364* 0.2^3*0.8^11?
dboyette
  • dboyette
the answer for 3
anonymous
  • anonymous
yes i think i didn't compute it so let me check
anonymous
  • anonymous
yes.
dboyette
  • dboyette
the # 5 would be 18018*0.2^5*0.8^9?
anonymous
  • anonymous
\[\dbinom{14}{3}=\frac{14\times 13\times 12}{3\times 2}=14\times 13\times 2=364\]
anonymous
  • anonymous
yes you have it
dboyette
  • dboyette
omg i think i am finally starting to get this, thank you so much
anonymous
  • anonymous
i can write a general formula for you if you like
dboyette
  • dboyette
please
anonymous
  • 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
  • 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
  • dboyette
ok
anonymous
  • 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
  • anonymous
in this case we had 14 people, in the general case we will have n people (tosses of a coin, whatever)
anonymous
  • 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
  • dboyette
yes
anonymous
  • 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
  • anonymous
that is like our \[.2^5\times .8^9\]
anonymous
  • 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
  • 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
  • anonymous
we computed this with n = 14, p = .2 and 1-p=.8 for various different values of k
dboyette
  • dboyette
so k was the different numbers we used for side effects
anonymous
  • anonymous
yes, the number of people who could have it. question one was for k = 4
anonymous
  • 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
  • anonymous
because "5 or more" means "not none, not one, not two, not three and not four"
dboyette
  • dboyette
what if they don't equal 1?
anonymous
  • anonymous
now sure what you are asking.
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
you can compute all these if you want, and add them and that will be your answer.
anonymous
  • 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
  • anonymous
this way you only have to compute 5 probabilities instead of 10
dboyette
  • dboyette
oh i see. that makes sense
anonymous
  • anonymous
good!
anonymous
  • 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
  • dboyette
thank you you are awesome!
anonymous
  • anonymous
welcome!

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