Respuesta :
Let [tex]L_1[/tex] denote the event that a student is a freshman, [tex]L_2[/tex] a sophomore, [tex]L_3[/tex] a junior, and [tex]L_4[/tex] a senior.
Let [tex]E[/tex] denote the event that a student majors in engineering. By the law of total probability,
[tex]\mathbb P(E)=\mathbb P(E\cap L_1)+\mathbb P(E\cap L_2)+\mathbb P(E\cap L_3)+\mathbb P(E\cap L_4)[/tex]
By the definition of conditional probability, we can expand each of these intersection probabilities to get
[tex]\mathbb P(E)=\mathbb P(E\mid L_1)\mathbb P(L_1)+\mathbb P(E\mid L_2)\mathbb P(L_2)+\mathbb P(E\mid L_3)\mathbb P(L_3)+\mathbb P(E\mid L_4)\mathbb P(L_4)[/tex]
[tex]\mathbb P(E)=0.15\cdot0.3+0.2\cdot0.25+2(0.3\cdot0.45)=0.365[/tex]
Let [tex]E[/tex] denote the event that a student majors in engineering. By the law of total probability,
[tex]\mathbb P(E)=\mathbb P(E\cap L_1)+\mathbb P(E\cap L_2)+\mathbb P(E\cap L_3)+\mathbb P(E\cap L_4)[/tex]
By the definition of conditional probability, we can expand each of these intersection probabilities to get
[tex]\mathbb P(E)=\mathbb P(E\mid L_1)\mathbb P(L_1)+\mathbb P(E\mid L_2)\mathbb P(L_2)+\mathbb P(E\mid L_3)\mathbb P(L_3)+\mathbb P(E\mid L_4)\mathbb P(L_4)[/tex]
[tex]\mathbb P(E)=0.15\cdot0.3+0.2\cdot0.25+2(0.3\cdot0.45)=0.365[/tex]
Probability of an event is the measure of its chance of occurrence. The probability that a random student at this college majors in engineering is 0.24
What is chain rule in probability?
For two events A and B, by chain rule, we have:
[tex]P(A \cap B) = P(B)P(A|B) = P(A)P(A|B)[/tex]
What is law of total probability?
Suppose that the sample space is divided in n mutual exclusive and exhaustive events tagged as
[tex]B_i \: ; i \in \{1,2,3.., n\}[/tex]
Then, suppose there is event A in sample space.
Then probability of A's occurrence can be given as
[tex]P(A) = \sum_{i=1}^n P(A \cap B_i)[/tex]
Using the chain rule, we get
[tex]P(A) = \sum_{i=1}^n P(A \cap B_i) = \sum_{i=1}^n P(A)P(B_i|A) = \sum_{i=1}^nP(B_i)P(A|B_i)[/tex]
For the given case, let we consider events:
- A = Student randomly selected is freshman
- B = Student randomly selected is sophomore
- C = Student randomly selected is junior or senior
- D = Student randomly selected majors in engineering.
Then, we have:
- P(A) = 0.3
- P(B) = 0.25
- P(C) = 0.45
- P(D|A) = 0.15 (Since A had happened already(as they said student is known to be a freshmen) and then we were calculating D's probability. Its notation is P(D|A) )
- P(D|B) = 0.2
- P(D|C) = 0.3
The needed probability is
P(random student at this college majors in engineering) = P(D)
Using the law of total probability, we get:
P(D) = P(A).P(D|A) + P(B).P(D|B) + P(C).P(D|C)
P(D) = 0.045 + 0.06 + 0.135 = 0.24
Thus,
The probability that a random student at this college majors in engineering is 0.24
Learn more about law of total probability here:
https://brainly.com/question/13225049
