Answer to Question #204958 in Statistics and Probability for DEBANKUR BISWAS

Question

1) An athlete is running in 5 races and in each race he has a 70% chance of winning. what is the probability that he will win at least two races?

2) the average number of cars arriving at a particular red light each day is 4. Assuming a poison distribution, calculate the probability that on a given day, less than three cars will arrive at the red light.

Answer 1

1) Let $X=$ the number of winned races: $X\sim Bin(n, p)$

Given $n=5, p=0.7$

$$P(X\geq2)=1-P(X=0)-P(X=1)$$

$$=1-\dbinom{5}{0}(0.7)^0(1-0.7)^{5-0}-\dbinom{5}{1}(0.7)^1(1-0.7)^{5-1}$$

$$=1-(0.3)^5-5(0.7)(0.3)^4$$

$$=1-3.8(0.3)^4=0.96922$$

The probability that he will win at least two races is 0.96922.

2) Let $X=$ the number of cars arriving at a particular red light : $X\sim Po(\lambda).$

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

Given $\lambda=4.$

$$P(X<3)=P(X=0)+P(X=1)+P(X=2)$$

$$=\dfrac{e^{-4}\cdot4^0}{0!}+\dfrac{e^{-4}\cdot4^1}{1!}+\dfrac{e^{-4}\cdot4^2}{2!}$$

$$=e^{-4}(1+4+8)\approx0.238103$$