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Mathematics 20 Online
OpenStudy (tomfoolery1):

A car insurance company has determined that 9% of all drivers were involved in a car accident last year. Among the 14 drivers living on one particular street, 3 were involved in a car accident last year. If 14 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year? 0.906 0.094 0.126 0.393

OpenStudy (tomfoolery1):

i tried the binomial distribution but now im just confused on where to start

OpenStudy (will.h):

can you tell me what you have so far

OpenStudy (tomfoolery1):

Well i though i could use b(x; n, P) – nCx * Px * (1 – P)n – x but i didnt work for me XD

OpenStudy (phi):

can you figure out the probability of getting 0 of 14 who had an accident using the binomial distribution?

OpenStudy (phi):

chance of getting 0 accidents : use n=14 x=0 P= 9% = 0.09 (1-P) = 0.91

OpenStudy (phi):

If you can, then you would find the chance of getting 0 , 1 or 2 accidents out of 14 add up those 3 numbers then 1- sum is the chance of getting 3 or more accidents.

OpenStudy (tomfoolery1):

do i take (14)(0.09)+0.91?

OpenStudy (phi):

isn't the formula \[n C x \cdot P^x \cdot (1-P)^{n-x} \] n choose x or in this case 14 C 0 is 1 (by definition, or use a calculator) P^0 is 1 so you are left with \[ 1 \cdot 1 \cdot (1-0.09)^{14} \]

OpenStudy (tomfoolery1):

i got 0.26704. so do id do that three times and add them together ?

OpenStudy (phi):

yes, that looks good. now find the probability of getting 1 accident out of 14 you do 14C1 * 0.09^1 * (1-0.09)^13 then for 2 accidents 14C2 * 0.09^2 * (1-0.09)^12 now add up the 3 numbers and finally to find the probability of (not 0,1 or 2) (i.e. for 3 or more) do 1- sum

OpenStudy (tomfoolery1):

0.267+0.238+0.370 = 0.875

OpenStudy (tomfoolery1):

1-0.875=0.125?

OpenStudy (phi):

your 2nd two numbers are off. Let's do x=1 14C1 * 0.09^1 * (1-0.09)^13 what do you get for 14 choose 1 ? and 0.09 to the 1st power is just 0.09 and what do you get for 0.91^13 ?

OpenStudy (tomfoolery1):

14 choose 1 is 14 . and 0.91^13 is 0.293

OpenStudy (phi):

ok, so you do 14*0.09*0.29345 (I would keep a few more decimals, and round at the end)

OpenStudy (tomfoolery1):

i got 0.370 like the last time

OpenStudy (phi):

oh, I didn't notice you did them in a different order. yes, then your numbers are ok (but you should have kept more decimals)

OpenStudy (phi):

and after you do 1-0.875=0.125 that is close to the actual choice. if you kept more decimals (at least 4) then the final answer , when rounded, would be 0.126

OpenStudy (tomfoolery1):

Ahhh thank you so much i see what i did XD

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