In Exercise 13, Ai={ awarded project i} , for I =1,2,3. Use the probabilities given there to compute the following probabilities, and explain in words the meaning of each one.
a. P(A2|A1)
b.P(A2 ∩A3|A1)
c.P(A2 ∪A3|A1)
d.P(A1 ∩A2 ∩A3|A1 ∪A2 ∪A3)
From the given information,
Let \(A_{i}=\{\) Awarded project \(i\}\), for \(i=1,2,3 .\) \(P\left(A_{1}\right)=0.22\) \(P\left(A_{2}\right)=0.25\) \(P\left(A_{3}\right)=0.28\) \(P\left(A_{1} \cap A_{2}\right)=0.11\) \(P\left(A_{1} \cap A_{3}\right)=0.05\) \(P\left(A_{2} \cap A_{3}\right)=0.07\) \(P\left(A_{1} \cap A_{2} \cap A_{3}\right)=0.01\)
(a)
Compute \(P\left(A_{2} \mid A_{1}\right)\)
$$ \begin{aligned} P\left(A_{2} \mid A_{1}\right) &=\frac{P\left(A_{1} \cap A_{2}\right)}{P\left(A_{1}\right)} \\ &=\frac{0.11}{0.22} \\ &=0.5 \end{aligned} $$
Therefore, the probability that projects 2 is awarded given that project 1 is awarded is 0.5.
(b)
Compute \(P\left(A_{2} \cap A_{3} \mid A_{1}\right)\)
$$ \begin{aligned} P\left(A_{2} \cap A_{3} \mid A_{1}\right) &=\frac{P\left(A_{2} \cap A_{3} \cap A_{1}\right)}{P\left(A_{1}\right)} \\ &=\frac{0.01}{0.22} \\ &=0.045 \end{aligned} $$
Therefore, the probability that projects 2 and 3 are awarded given that project 1 is awarded is 0.045.
(c)
Compute \(P\left(A_{2} \cup A_{3} \mid A_{1}\right)\).
$$ \begin{aligned} P\left(A_{2} \cup A_{3} \mid A_{1}\right) &=\frac{P\left(\left(A_{2} \cup A_{3}\right) \cap A_{1}\right)}{P\left(A_{1}\right)} \\ &=\frac{P\left(\left(A_{2} \cap A_{1}\right) \cup\left(A_{3} \cap A_{1}\right)\right)}{P\left(A_{1}\right)} \\ &=\frac{P\left(A_{2} \cap A_{1}\right)+P\left(A_{3} \cap A_{1}\right)-P\left(A_{1} \cap A_{2} \cap A_{3}\right)}{P\left(A_{1}\right)} \\ &=\frac{0.11+0.05-0.01}{0.22} \\ &=0.682 \end{aligned} $$
Therefore, the probability that projects 2 or 3 are awarded given that project 1 is awarded is 0.682.
(d)
Compute \(P\left(A_{1} \cap A_{2} \cap A_{3} \mid A_{1} \cup A_{2} \cup A_{3}\right)\).
$$ \begin{aligned} P\left(A_{1} \cap A_{2} \cap A_{3} \mid A_{1} \cup A_{2} \cup A_{3}\right) &=\frac{P\left(\left(A_{1} \cap A_{2} \cap A_{3}\right) \cap\left(A_{1} \cup A_{2} \cup A_{3}\right)\right)}{P\left(A_{1} \cup A_{2} \cup A_{3}\right)} \\ &=\frac{P\left(A_{1} \cap A_{2} \cap A_{3}\right)}{P\left(A_{1} \cup A_{2} \cup A_{3}\right)} \end{aligned} $$
Here,
$$ \begin{aligned} P\left(A_{1} \cup A_{2} \cup A_{3}\right)=& P\left(A_{1}\right)+P\left(A_{2}\right)+P\left(A_{3}\right)-P\left(A_{1} \cap A_{2}\right)-P\left(A_{1} \cap A_{3}\right) \\ &-P\left(A_{2} \cap A_{3}\right)+P\left(A_{1} \cap A_{2} \cap A_{3}\right) \\ =& 0.22+0.25+0.28-0.11-0.05-0.07+0.01 \\ =& 0.53 \end{aligned} $$
Substitute the value in the above formula.
$$ \begin{aligned} P\left(A_{1} \cap A_{2} \cap A_{3} \mid A_{1} \cup A_{2} \cup A_{3}\right) &=\frac{P\left(A_{1} \cap A_{2} \cap A_{3}\right)}{P\left(A_{1} \cup A_{2} \cup A_{3}\right)} \\ &=\frac{0.01}{0.53} \\ &=0.019 \end{aligned} $$
Therefore, the probability that all three projects are awarded, given that at least one of them is awarded is 0.019.