Question

if the probability that an airplane arrivea on time is 0.75 find prob that 1a) out of three flights all of them arrive on time. 1B) out of 3 flights at least one of them arrives on time?

Answer 1

To answer this questions we can use the binomial distribution.

The binomial distribution is given by:

$P(X=k)=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$

where n is the number of experiments, k is the number of times of success of the experiment and p is the probability of succes.

Remember that the factorial of an integer number is defined as:

$n!=n(n-1)(n-2)\cdot\cdot\cdot\cdot\cdot3\cdot2\cdot1$

For example:

$\begin{gathered} 3!=3\cdot2\cdot1=6 \\ 2!=2\cdot1=2 \\ 1!=1 \\ 0!=1 \end{gathered}$

1a.

In this case we are gonna do the experiment 3 times (that is we are going to analyze three flights). We want to know the probability of all the flights to be on time, this means that we want that k=3. Furthermore we know that the probability of the flight arriving on time is 0.75, then p=0.75. Plugging this values into the formula we have:

$\begin{gathered} P(X=3)=\frac{3!}{3!(3-3)!}0.75^3(1-0.75)^{3-3} \\ =0.421875 \end{gathered}$

Therefore the probability of three out of three flights arriving on time is 0.421875

1b.

In this case we want the following probability:

$P(X\ge1)$

that is, we want to find the probability of at least one flight arriving on time.

To find this probability we can do it in two ways:

$P(X\ge1)=P(X=1)+P(X=2)+P(X=3)$

or, we can do it if we remember that the sum of all possible porbabilities is equal to one, then:

$P(X\ge1)=1-P(X=0)$

The second form is a little easier to calculate so we are going to use that one. In that case we have that:

$\begin{gathered} P(X\ge1)=1-\frac{3!}{0!(3-0)!}(0.75)^0(1-0.75)^{3-0} \\ =1-\text{0}.015625 \\ =0.984375 \end{gathered}$

Therefore the probability of at least one out of three flights arriving on time is 0.984375