probability question

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GabeFried

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Suppose that 83% of the families in a town own a cellphone. If 10 families are surveyed at random, a) what is the probability that exactly four families own a cellphone? B) what is the probability that at least seven families own a cellphone?

for A) heres what i have so far (0.83)^4 x (0.17)^10-4
when solving im getting 1.14 which is wrong.
p(x=4)
n=10


for B i know that i would have to do the same as A as an example P7 + P8 + P9 +P10
(0.83)^4 x (0.17)^10-4 + (0.83)^4 x (0.17)^10-4 + (0.83)^4 x (0.17)^10-4


i know thats wrong but i want to get A first and then i should be able to do B myself
 
.83 is less than 1. So if you multiply it by itself 4 times should it be less than 1, bigger than 1 or equal to 1? Do NOT use a calculus to decide.
.17 is less than 1. So if you multiply it by itself 6 times should it be less than 1, bigger than 1 or equal to 1? Do NOT use a calculus to decide.
Does it make sense that (0.83)^4 x (0.17)^6 is 1.14??

Part b. You should know that if you add (0.83)^4 x (0.17)^6 to itself 3 times that will get 3(0.83)^4 x (0.17)^6. Why are you adding it 3 times? Where did you get (0.83)^4 x (0.17)^6? What does it mean at least 7 mean in this problem?
 
0.84^4 is 0.4745....
0.17^6 = 2.41
so i still think i have something wrong as the result is 1.2017


for b i said that is just and example of how i plan to do it that isnt what im actually going to put as the final answer
 
GabeFried said:
Suppose that 83% of the families in a town own a cellphone. If 10 families are surveyed at random, a) what is the probability that exactly four families own a cellphone? B) what is the probability that at least seven families own a cellphone?

for A) heres what i have so far (0.83)^4 x (0.17)^10-4
when solving im getting 1.14 which is wrong.
p(x=4)
n=10

for B i know that i would have to do the same as A as an example P7 + P8 + P9 +P10
(0.83)^4 x (0.17)^10-4 + (0.83)^4 x (0.17)^10-4 + (0.83)^4 x (0.17)^10-4

i know thats wrong but i want to get A first and then i should be able to do B myself
Click to expand...
You're omitting an essential part of the binomial probability formula. Isn't there supposed to be a binomial coefficient (combinations) in there somewhere?

GabeFried said:
0.84^4 is 0.4745....
0.17^6 = 2.41
so i still think i have something wrong as the result is 1.2017
Click to expand...

As for the calculation itself,

0.83^4 = 0.47458321​
0.17^6 = 0.000024137569​

It looks like you missed the "E-5" on your calculator's display, which means "x10^-5".
 
.17 = 17%. If you take 17% of a number it gets smaller!

.17*.17 = 17%*.17 (ie 17% of .17) which is less that .17 which is less than 1.

You can not multiply positive numbers less than 1 and get bigger than 1.

Stop believing whatever your calculator tells you!

Now please try again and compute 0.17^6. Also your result for 0.84^4 is wrong as well.
 
Dr.Peterson said:
You're omitting an essential part of the binomial probability formula. Isn't there supposed to be a binomial coefficient (combinations) in there somewhere?



As for the calculation itself,

0.83^4 = 0.47458321​
0.17^6 = 0.000024137569​

ah yes, so 10C4 x (0.000024137569 x 0.47458321) = 0.606213681 x100 = 60.6%?
Click to expand...
 
Jomo said:
.17 = 17%. If you take 17% of a number it gets smaller!

.17*.17 = 17%*.17 (ie 17% of .17) which is less that .17 which is less than 1.

You can not multiply positive numbers less than 1 and get bigger than 1.

Stop believing whatever your calculator tells you!

Now please try again and compute 0.17^6. Also your result for 0.84^4 is wrong as well.
Click to expand...
its 83% in the question and when i calculate 83^4 i keep getting 0.47458321 so im unsure what you mean and yes for the 01.7^6 i missed the scientific notation 0.000024137569 is that answer
 
It is best to leave your answers as a number between 0 and 1 inclusive and not as a percent.
 
for part B i have
10C7 x (.83)^7X(0.17)3 + 10C8 x (.83)^8X(0.17)^2 + 10C9 x (.83)^9X(0.17)^1 +10C10 x (.83)^10X(0.17)^0

unsure if i have the combination values are correct
 
GabeFried said:
Suppose that 83% of the families in a town own a cellphone. If 10 families are surveyed at random, a) what is the probability that exactly four families own a cellphone? B) what is the probability that at least seven families own a cellphone?
Click to expand...
Please look at this LINK for part A. LINK for part B.
To:GabeFried Looking over your posts, one must wonder if you have even a rudimentary idea what is going on with probability.
Please study those links. Use that website to do your caculations.
 
GabeFried said:
ah yes, so 10C4 x (0.000024137569 x 0.47458321) = 0.606213681 x100 = 60.6%?
Click to expand...
Not quite; try that again. I don't get anything near 0.606213681. Maybe show what you get for each piece, so we can see where things go wrong in your calculator usage.
 
Dr.Peterson said:
Not quite; try that again. I don't get anything near 0.606213681. Maybe show what you get for each piece, so we can see where things go wrong in your calculator usage.
Click to expand...
i redid it and got 0.00240.....that was my bad i appreciate the nudge in the right direction
 
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