Question 1096463
<font color="black" face="times" size="3">
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 <font color=red size=4>0.1963</font> (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%
</font>