Respuesta :
Answer:
0.289792.
Step-by-step explanation:
If we define a month with one or more accident as "success"; and
A month with no accident as "failure" .
- P(one or more accidents will occur during any given month)=0.60.
- P(no accident will occur during any given month)=1-0.60=0.40
We want to calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs.
in this case, let k=number of failures before the fourth success
Therefore, [tex]k\geq 4[/tex]
k follows a negative binomial distribution.
Therefore, the probability is:
[tex]P(k\geq 4)=1-P(k\leq 3)\\=1-\sum_{k=0}^{3}\left(\begin{array}{ccc}3+k\\k\end{array}\right)(0.60)^4(0.40)^k\\=1-(0.60)^4 \left[\left(\begin{array}{ccc}3\\0\end{array}\right)(0.40)^0+\left(\begin{array}{ccc}4\\1\end{array}\right)(0.40)^1+\left(\begin{array}{ccc}5\\2\end{array}\right)(0.40)^2+\left(\begin{array}{ccc}6\\3\end{array}\right)(0.40)^3\right]\\=1-(0.60)^4 \left[ 1+1.6+1.6+1.28\right]\\=1-(0.60)^4[5.48]\\=0.289792[/tex]
The required probability is 0.289792.
