Question

1.     

          (ii) If 1% of an airline’s flight suffers a minor equipment failure in an aircraft, what is the probability that there will be exactly 5 such in the next 1000 such flights?                                (2)

     

Answer 1

$p=0.01, n=1000, q=1-p=0.99$

$$P(X=5)=\dbinom{1000}{5}(0.01)^5(0.99)^{100-5}=$$

$$=0.03745311161$$

The probability that there will be exactly 5 such in the next 1000 such flights is

$0.03745311161.$

Use the approximation of binomial distribution by Poisson

$$\lambda=np=0.01(1000)=10$$

$$P(X=5)=\dfrac{e^{-10}\cdot10^5}{5!}\approx0.0378332748$$