Problem

# In Exercise 13, Ai={ awarded project i} , for I =1,2,3. Use the probabilities given ther...

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)

#### Step-by-Step Solution

Solution 1

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.