Question 1096463: what is the answer for :
For drivers aged 20-24 there is a 34% chance of having a car accident in a one year period (based on data from the National Safety Council). In a sample of 12 drivers aged 20-24, let X be the number who have had an accident in the last year. Consider the following probability distribution for X. Find P(x), when x = 5
Given table
x P(x)
6 0.1180
7 0.0521
8 0.0168
9 0.0038
10 0.0006
11 0.0001
12 0.0000
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
This is a binomial probability distribution problem.
The sample size is n = 12 and we want k = 5 to be the number of people who had an accident.
The probability of getting into an accident, for any given independent trial, is p = 0.34 (34% = 34/100 = 0.34)
Compute the combination value below
n C k = (n!)/(k!*(n-k)!)
12 C 5 = (12!)/(5!*(12-5)!)
12 C 5 = (12!)/(5!*7!)
12 C 5 = (12*11*10*9*8*7!)/(5!*7!)
12 C 5 = (12*11*10*9*8)/(5!)
12 C 5 = (12*11*10*9*8)/(5*4*3*2*1)
12 C 5 = 95040/120
12 C 5 = 792
Use that result to compute the binomial distribution value
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 5) = (12 C 5)*(0.34)^(5)*(1-0.34)^(12-5)
P(X = 5) = (12 C 5)*(0.34)^(5)*(0.66)^(7)
P(X = 5) = (792)*(0.34)^(5)*(0.66)^7
P(X = 5) = (792)*(0.0045435424)*(0.05455160701056)
P(X = 5) = 0.196303171236968
Rounding to four decimal places, we get the final answer of 0.1963 (this answer is approximate)
This means that if we selected 12 people, then the chances that exactly five of them got into an accident last year is roughly 0.1963
Note: 0.1963 is equivalent to 19.63%
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