91Ó°ÊÓ

The data are provided that consists of pmf of a random variable Y:

a.

Here, the probability thatthe flight will accommodate all ticketed passengers who show up is obtained.

The required probability is\(P\left( {Y \le 50} \right)\). It can be computed as:

\(P\left( {Y \le 50} \right) = \left( \begin{array}{l}P\left( {X = 45} \right) + P\left( {X = 46} \right) + P\left( {X = 47} \right)\\ + P\left( {X = 48} \right) + P\left( {X = 49} \right) + P\left( {X = 50} \right)\end{array} \right)\)

Substitute the values of\(P\left( {X = 45} \right) = 0.05\),\(P\left( {X = 46} \right) = 0.1\),\(P\left( {X = 47} \right) = 0.12\),\(P\left( {X = 48} \right) = 0.14\),\(P\left( {X = 49} \right) = 0.25\)and\(P\left( {X = 50} \right) = 0.17\)in the above formula,

\(\begin{array}{c}P\left( {X \le 50} \right) = 0.05 + 0.1 + 0.12 + 0.14 + 0.25 + 0.17\\ = 0.83\end{array}\)

Thus, the required probability is 0.83.

b.

Here, the probability thatnot all ticketed passengers who show up can be accommodated is obtained.

So, the required probability is\(1 - P\left( {Y \le 50} \right)\). It can be computed as:

\(1 - P\left( {Y \le 50} \right) = 1 - \left( \begin{array}{l}P\left( {X = 45} \right) + P\left( {X = 46} \right) + P\left( {X = 47} \right)\\ + P\left( {X = 48} \right) + P\left( {X = 49} \right) + P\left( {X = 50} \right)\end{array} \right)\)

Substitute the values of\(P\left( {X = 45} \right) = 0.05\),\(P\left( {X = 46} \right) = 0.1\),\(P\left( {X = 47} \right) = 0.12\),\(P\left( {X = 48} \right) = 0.14\),\(P\left( {X = 49} \right) = 0.25\)and\(P\left( {X = 50} \right) = 0.17\)in the above formula,

\(\begin{array}{c}1 - P\left( {X \le 50} \right) = 1 - \left( {0.05 + 0.1 + 0.12 + 0.14 + 0.25 + 0.17} \right)\\ = 1 - 0.83\\ = 0.17\end{array}\)

Thus, the required probability is 0.17.

c.

In this case, if a person is the first person on the stand list then the probability that a person will able to take the flight is when there is one seat left out of 50 seats or at most 49 ticketed passengers would shows up in a plane.

Therequired probability is\(P\left( {Y \le 49} \right)\). It can be computed as:

\(P\left( {Y \le 49} \right) = \left( \begin{array}{l}P\left( {X = 45} \right) + P\left( {X = 46} \right) + P\left( {X = 47} \right)\\ + P\left( {X = 48} \right) + P\left( {X = 49} \right)\end{array} \right)\)

Substitute the values of\(P\left( {X = 45} \right) = 0.05\),\(P\left( {X = 46} \right) = 0.1\),\(P\left( {X = 47} \right) = 0.12\),\(P\left( {X = 48} \right) = 0.14\),\(P\left( {X = 49} \right) = 0.25\)and\(P\left( {X = 50} \right) = 0.17\)in the above formula,

\(\begin{array}{c}P\left( {X \le 49} \right) = \left( {0.05 + 0.1 + 0.12 + 0.14 + 0.25} \right)\\ = 0.66\end{array}\)

Thus, the required probability is 0.66.

In this case, if a person is the third person on the stand list then the probability that person will able to take the flight is when there is three seats left out of 50 seats or at most 47 ticketed passengers would shows up in a plane.

Therequired probability is\(P\left( {Y \le 47} \right)\). It can be computed as:

\(P\left( {Y \le 47} \right) = P\left( {X = 45} \right) + P\left( {X = 46} \right) + P\left( {X = 47} \right)\)

Substitute the values of\(P\left( {X = 45} \right) = 0.05\),\(P\left( {X = 46} \right) = 0.1\), and\(P\left( {X = 47} \right) = 0.12\),in the above formula,

\(\begin{array}{c}P\left( {X \le 47} \right) = \left( {0.05 + 0.1 + 0.12} \right)\\ = 0.27\end{array}\)

Thus, the required probability is 0.27.