Question

Five computer program modules are ranked as M1, M2, M3, M4, and M5 according to the ascending order of effort required to debug them, among which M1 requires the least amount of effort and M5 requires the most amount of effort. A software engineer must randomly select three program modules from the five modules. He (or she) does not know the amount of effort each module costs when selecting.
A) What is the sample space?
B) Let A be the event that one selected module requires the least amount of effort. List the outcomes in A, and calculate the probability of A.
C) Let B be the event that the one selected module requires the most effort. List the outcomes in B, and calculate the probability of B.
D) List the outcomes in the compound event AnB, and calculate the probability of it.
E) List the outcomes in the compound event AUB, and calculate the probability of it.
F) List the outcomes in the compound event AnB, and calculate the probability of it.
G) List the outcomes in the compound event AUB, and calculate the probability of it.
H) Are events A and B mutually exclusive? Explain why?

Answer 1

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Technician selects three out of 5 systems  

In C(5,3)=10ways, this can be achieved  

In part a:

Space sample chooses 3 of a 5 systems  

$(M_1, \ M_2,\ M_3),$$(M_1,M_2,M_4) \ (M_1,M_2,M_5) \ (M_1,M_3,M_4) \ (M_1,M_3,M_5),$$(M_1,M_4,M_5) \ (M_2,M_3,M_4)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5)}$

In point b:

A =MODULE WHICH INCLUDE M1 minimal amount of effort  

Outcomes probable =

$(M_1,M_2,M_3),\ (M_1,M_2,M_4) \ (M_1,M_2,M_5)\ (M_1,M_3,M_4)\\\\(M_1,M_3,M_5),\ (M_1,M_4,M)5)\ =\ 6$

$\to p(A)=\frac{6}{10}$

            $=0.6$

In point c:

B = highest effort that is $M_5$

Potential result=

$(M_1,M_2,M_5) \ (M_1,M_3,M_5) \ (M_2,M_3,M_5)\(M_2,M_4,M_5) \\ (M_2,M_4,M_5), \ (M_3,M_4,M_5) \ =\ 6$

$\to B= \frac{6}{10}$

        $=0.6$

$\to P(B)=10$

In point d:

$\to \ A \ intersection \ B=(M_1,M_2,M_5), \ (M_1,M_3,M_5) \ ,(M_1,M_4,M_5)$

$\to A (A \ intersection \ B) = \frac{3}{10}$

                                      $=0.3$

In point e:

$\to (A \cup B) = (M_1,M_2,M_3),\ (M_1,M_2,M_4)\ (M_1,M_2,M_5)(M_1,M_3,M_4)\ (M_1,M_3,M_5), \\ (M_1,M_4,M_5)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5) \ = \ 9$$\to P(A \cap B)=\frac{9}{10}$

                    $= 0.9$

In point f:

$\to (A\cap B) = \frac{3}{10}$

                 $= 0.3$

In point g:

$\to (A \cup B) = \frac{7}{10}$

                 $=0.7$

In point h:

$\to p(A \cap B) = 0.3 \neq 0$