Question

1.     What is the probability of getting a total of 7 or 11 when a pair of fair dice is tossed?

2.     In a certain industrial facility , accidents occur in frequently. It is known that the probability of an accident on any given day is 0.005 and accidents are independent of each other.

a)     What is the probability that in any given period of 400 days , there will be an accident on one day?

b)    What is the probability that in any given period of 400 days , there are at most three days with an accident?

Answer 1

1.

$$\def\arraystretch{1.5} \begin{array}{c:c:c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hdashline 2 & 3 & 4 & 5 & 6 & 7 & 8\\ \hdashline 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hdashline 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hdashline 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hdashline 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hdashline \end{array}$$

$$P(7\ or 11)=\dfrac{6+2}{36}=\dfrac{2}{9}$$

2.

a) Note that $n=400$ and $p=0.005$.

Using Poisson approximation with $\lambda=np=400(0.005)=2$ we have

$$P(X=1)=\dfrac{e^{-\lambda}\cdot\lambda^1}{1!}=\dfrac{e^{-2}\cdot2^1}{1!}$$

$$=2e^{-2}\approx0.27067$$

b) Note that $n=400$ and $p=0.005$.

Using Poisson approximation with $\lambda=np=400(0.005)=2$ we have

$$P(X\leq 3)=P(X=0)+P(X=1)$$

$$+P(X=2)+P(X=3)=$$

$$=\dfrac{e^{-2}\cdot2^0}{0!}+\dfrac{e^{-2}\cdot2^1}{1!}+\dfrac{e^{-2}\cdot2^2}{2!}+\dfrac{e^{-2}\cdot2^3}{3!}$$

$$=\dfrac{19}{3}\cdot e^{-2}\approx0.85712$$