91Ó°ÊÓ

Simply put, probability refers to the likelihood of something occurring. We can discuss the probabilities of certain outcomes—how likely they are—when we're unsure about the outcome of an event. Statistics refers to the study of events that are governed by chance.

We have events

\(\begin{aligned}{l}{A_i} &= \{ {\rm{ awarded project }}i\} \;\;\\i &= 1,2,3,\end{aligned}\)

and also, the following probabilities

\(\begin{aligned}{l}P\left( {{A_1}} \right) = 0.22;\\P\left( {{A_2}} \right) = 0.25;\\P\left( {{A_3}} \right) = 0.28\\P\left( {{A_1} \cap {A_2}} \right) = 0.11;\\P\left( {{A_1} \cap {A_3}} \right) = 0.05;\\P\left( {{A_2} \cap {A_3}} \right) = 0.07;\\P\left( {{A_1} \cap {A_2} \cap {A_3}} \right) = 0.01;\end{aligned}\)

We need to find the conditional probability of \({A_2}\)given that the event \({A_1}\)has occurred, which means that we need to find the probability that project \(2\) was awarded if we know that project \(1\) was awarded.

The conditional probability of A given that the event B has occurred, for which\(P(B) > 0\), is

The conditional probability of A given that the event B has occurred, for which \(P(B) > 0\), is

\(P(A\mid B) = \frac{{P(A \cap B)}}{{P(B)}}\)

for any two events A and B.

Using the definition, we have

\(P\left( {{A_2}\mid {A_1}} \right) = \frac{{P\left( {{A_1} \cap {A_2}} \right)}}{{P\left( {{A_1}} \right)}}\mathop = \limits^{(1)} \frac{{0.11}}{{0.22}} = 0.5,\)

(1): given in the exercise, see above.

We need to find conditional probability of \({A_2} \cap {A_3}\) given that the event \({A_1}\)has occurred, which means that we need to find probability that the project \(2\) and \(3\) were awarded if we know that project \(1\) was awarded.

From the definition of conditional probability given above we have

\(P\left( {{A_2} \cap {A_3}\mid {A_1}} \right) = \frac{{P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)}}{{P\left( {{A_1}} \right)}} = \frac{{0.01}}{{0.22}} = 0.0455,\)

where all the probabilities are given (see above).

We need to find conditional probability of \({A_2} \cup {A_3}\)given that the event \({A_1}\)has occurred, which means that we need to find probability that the project \(2\) or project \(3\)or both were awarded if we know that project \(1\) was awarded.

From the definition of conditional probability given above we have

\(\begin{aligned}P\left( {{A_2} \cup {A_3}\mid {A_1}} \right) &= \frac{{P\left( {\left( {{A_1} \cup {A_2}} \right)\cap{A_3}} \right)}}{{P\left( {{A_1}} \right)}} \hfill \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= \frac{{P\left( {\left( {{A_1}\cap {A_2}} \right) \cup \left( {{A_1} \cap {A_3}} \right)} \right)}}{{P\left( {{A_1}} \right)}} \hfill \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= \frac{{P\left( {{A_1} \cap {A_2}} \right) + P\left( {{A_1} \cap {A_3}} \right) - P\left({{A_1} \cap {A_2} \cap {A_3}} \right)}}{{P\left( {{A_1}} \right)}} \hfill \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &=\frac{{0.11 + 0.05 - 0.01}}{{0.22}} \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= \frac{{0.15}}{{0.22}} = 0.682, \hfill \\\end{aligned} \)

(1): this stands for any three events,

(2): here we use the following proposition for \(A = {A_1} \cap {A_2}\)and \(B = {A_1} \cap {A_3}\):

Proposition: For every two events A and B

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

We need to find conditional probability of \({A_1} \cap {A_2} \cap {A_3}\)given that the event \({A_1} \cup {A_2} \cup {A_3}\) has occurred, which means that we need to find probability that the project \(1\) and the project \(2\) and project \(3\) were awarded if we know that project \(1\) or project \(2\) or project \(3\) or any combination of the three events was awarded (in other words: at least one project was awarded).

From the definition of conditional probability given above we have

\(\begin{aligned}P\left( {{A_1} \cap {A_2} \cap {A_3}\mid {A_1} \cup {A_2} \cup {A_3}} \right) &=\frac{{P\left({\left( {{A_1} \cap {A_2} \cap {A_3}} \right) \cap \left( {{A_1} \cup {A_2} \cup {A_3}} \right)}\right)}}{{P\left( {{A_1} \cup {A_2} \cup {A_3}} \right)}} \hfill \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\,\,\;\;\;\;\;\; &=\frac{{P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)}}{{\left. {P\left( {{A_1} \cup {A_2} \cup{A_3}} \right)} \right)}} \hfill \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &=\frac{{0.01}}{{0.53}} = 0.0189, \hfill \\\end{aligned}\)

(1): the intersection is subset of the union, and when\(A \subseteq B\;then\;A \cap B = A\),

(2): we compute the probability of the union as follows

\(\begin{aligned}P\left( {{A_1} \cup {A_2} \cup {A_3}} \right) &= P\left( {{A_1}} \right) + P\left({{A_2}} \right) + P\left( {{A_3}} \right) - P\left( {{A_1} \cap {A_2}} \right) - P\left({{A_1} \cap {A_3}} \right) - P\left( {{A_2} \cap {A_3}} \right) + P\left( {{A_1} \cap {A_2}\cap {A_3}} \right) \hfill \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= 0.22 + 0.25 + 0.28 + 0.11 - 0.05 -0.07 + 0.01 \hfill \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= 0.53 \hfill \\\end{aligned} \)

Here, we use the proposition that stands for any three events.

Proposition: For every three events A, B and C the following is true

\(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\)