Respuesta :
Answer:
a) The Venn diagram is presented in the attached image to this answer.
b) Check Explanation.
c) 0.3333
d) 0.1429
e) 0.6
Explanation:
Let the probability of a lake having high BOD be P(B) = 30% = 0.3
Probability of a lake having low pH = P(P) = 20% = 0.2
Probability that a lake has high BOD and low pH = P(B n P) = 10% = 0.1
Then, probability that a lake has normal BOD = P(B') = 1 - P(B) = 1 - 0.3 = 0.7
Probability that a lake has normal pH = P(P') = 1 - P(P) = 1 - 0.2 = 0.8
Total probability = P(U) = 100% = 1
a) The Venn diagram is presented in the attached image to this answer.
b) Two events are independent if and only if, P(A|B) = P(A) or P(B|A) = P(B).
For this question,
P(B|P) = P(B n P)/P(P) = 0.1/0.2 = 0.5 ≠ P(B) (which is 0.3)
And P(P|B) = P(B n P)/P(B) = 0.1/0.3 = 0.333 ≠ P(P) (which is 0.2).
It is evident that the two events aren't independent of each other.
c) If a stream has high BOD, what is the probability it will also have low pH?
This probability is given as P(P|B) meaning that, the probability of a lake having low pH given that it has high BOD.
Mathematically, this conditional probability is given by
P(P|B) = P(B n P)/P(B) = 0.1/0.3 = (1/3) = 0.3333
d) If a stream has normal levels of BOD, what is the probability it will also have low pH.
This probability is given as P(P|B'); that is, the probability of a lake having low pH given that it has normal BOD.
Mathematically,
P(P|B') = P(B' n P)/P(B')
P(B') = 0.7 (already found above)
But P(B' n P) = ?
Mathematically,
P(B' n P) = P(P) - P(B n P) = 0.2 - 0.1 = 0.1
P(P|B') = 0.1/0.7 = 0.1429
e) What is the probability that a stream will not exhibit either pollution indicator, i.e., will have normal BOD and pH levels?
This is given as P(B' n P')
Mathematically, this represents the region in the Venn diagram outside of the circles representing P(B) and P(P) and it's given mathematically as,
P(B' n P') = P(U) - [P(B n P') + P(B' n P) + P(B n P)] = 1 - (0.2 + 0.1 + 0.1) = 1 - 0.4 = 0.6 or 60%
When the stream has high BOD the probability of low pH is 0.33 and 0.142 when it has normal BOD. 0.6 is the probability that a stream will exhibit neither indicator.
What is BOD?
BOD is the biological oxidation demand, that tells about the dissolved oxygen amount in the water body. BOD along with pH are the indicator of pollution.
The Venn diagram is attached in the image below.
The two indicators, high BOD and low pH are dependent on each other. It can be shown as:
[tex]\rm P(A|B) = P(A) \;or \;P(B|A) = P(B)[/tex]
But,
[tex]\begin{aligned}\rm P(B|P) &= \rm \dfrac{P(B \;n \;P)}{P(P)} \\\\&= \dfrac{0.1}{0.2} \\\\&= 0.5 \neq \rm P(B)\end{aligned}[/tex]
And
[tex]\begin{aligned}\rm P(P|B) &=\rm \dfrac{P(B \;n \;P)}{P(B)} \\\\&= \dfrac{0.1}{0.3} \\\\&= 0.333 \neq \rm P(P) \end{aligned}[/tex]
Hence they are not independent of each other.
The probability of low pH at high BOD is given as P(P|B).
[tex]\begin{aligned}\rm P(P|B) &= \rm \dfrac{P(B \;n \;P)}{P(B)}\\\\ &= \dfrac{0.1}{0.3}\\\\&= 0.3333\end{aligned}[/tex]
Hence, 0.33 is the probability of low pH at high BOD.
The probability of low pH at normal BOD is given as P(P|B').
[tex]\begin{aligned}\rm P(P|B') &= \rm \dfrac{P(B' \;n\; P)}{P(B')}\\\\\rm P(P|B') &= \dfrac{0.1}{0.7} \\\\&= 0.1429\end{aligned}[/tex]
Hence, 0.1429 is the probability of low pH at normal BOD.
The probability that a stream will not exhibit any of the indicators is given by, P(B' n P').
[tex]\begin{aligned}\rm P(B' n P') &= \rm P(U) - [P(B n P') + P(B' n P) + P(B n P)]\\\\&= 1 - (0.2 + 0.1 + 0.1) \\\\&= 0.6\end{aligned}[/tex]
Hence, 0.6 is the probability that neither of the indicators will be expressed.
Learn more about BOD and low pH here:
https://brainly.com/question/9866894
