specificagent1
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Is this part of our syllabus? (1 Viewer)
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Since this question appeared in a Mathematics Advanced 2020 trial paper(s), it is part of the syllabus. The working for this question is as follows:
I hope this helps!
I hope this helps!
specificagent1
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what paper is this? also i did not know this content existed... so screwed. Dont really understand all the notations and the A' stuff, i just do probability by thinking (if that makes sense?)Since this question appeared in a Mathematics Advanced 2020 trial paper(s), it is part of the syllabus. The working for this question is as follows:
View attachment 32965
I hope this helps!
This is from the Hunters Hill High School Mathematics Advanced 2020 trial paper.
Since this is a multiple-choice question, your working does not matter as long as you get the correct answer.
The apostrophe is used to denote the complement of a set. A' is all the items that are not in set A.
specificagent1
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what does the | mean. So what does P(B|A') geeze can do calculus but not this... rip me
5uckerberg
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| means what is the probability of something given that the previous event has occurred. Eg, P(B|A) is the probability of B given A. Independent result is when A=A' in terms of probability where A' refers to the complement of the set of A, and usually A=p and A'=1-p which is the complement where p refers to probability, but if they are equal then p=0.5 and this is an even chance which implies results are independent.
=\frac{P(A|B)P(B)}{P(A)}=m)
and =\frac{P(A'|B)P(B)}{P(A')}=n)
The independent part is very apparent because the formula can be written like this
I wanted to ask have I made a mistake somewhere?
The independent part is very apparent because the formula can be written like this
I wanted to ask have I made a mistake somewhere?
Last edited:
specificagent1
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ummm could you dumb that down please. I dont really get it
As was mentioned above, the vertical line "|" means "given that". This means that in the case of P(B | A'), the probability of B depends on that of A', meaning you must be mindful of this in your calculations in order to provide a correct answer.
A more detailed way of solving this question is as follows:
Here, the following formula is being used:
@ specificagent it look like Jimmy has answered
Just couple points
- Make sure to remember the definition of conditional probability
- What is the relationship between logic and set theory operators
- When do you use bayes theorem over the definition of conditional probability
Another approach...
Consider a tree diagram:
The question already gives us that = p \qquad \implies \qquad P(\text{not}\ A) = P(A') = 1 - p)
.
Following on, let = q \qquad \implies \qquad P(\text{not}\ B) = P(B') = 1 - q)
, and note that )
.
The rule for conditional probability is that
 = P(X|Y) = \cfrac{P(X \cap Y)}{P(Y)} \qquad \qquad \text{. . . . . . . . . . (1)})
which we need to use interpret the information that has been provided.
We seek the condition for events
and 
to be independent, which requires that
 = P(A) \times P(B) \qquad \qquad \text{. . . . . . . . . . (2)})
Now, our first piece of information is that = m)
:
 and using $X = B$ and $Y = A$, we get:} \quad P(B \cap A) &= P(B|A) \times P(A) \\ &= m \times p \\ P(A \cap B) &= mp \qquad \text{as $P(A \cap B) = P(B \cap A)$} \\ \text{And, applying (2):} \quad p \times q &= mp \\ q &= m \qquad \text{. . . . . . . . . . (3)} \end{align*})
And, our second piece of information is that = n)
:
 and using $X = B$ and $Y = A'$, we get:} \quad P(B \cap A') &= P(B|A') \times P(A') \\ &= n \times (1 - p) \\ P(A' \cap B) &= n(1 - p) \qquad \text{as $P(A' \cap B) = P(B \cap A')$} \\ \text{And, applying (2) as $P(A'\cap B) = P(A') \times P(B)$:} \quad (1 - p) \times q &= n(1 - p) \\ q &= n \\ m &= n \qquad \text{using (3)} \end{align*})
Consider a tree diagram:
The question already gives us that
Following on, let
The rule for conditional probability is that
which we need to use interpret the information that has been provided.
We seek the condition for events
Now, our first piece of information is that
And, our second piece of information is that
@specificagent1 there haven't been that many questions asked on independence, but they've mostly been relative challenging, FYI.
Some other conditional probability / independence questions
Girraween 2020 q3
Manly Selective 2020 q3, 19
Sydney Girls 2020 q7, 36
Killara 2020 q10
Normanhurst Boys 2020 q10, 20
Sydney Boys 2020 q28
Sydney Grammar 2020 q28
North Sydney Girls 2020 q29
North Sydney Boys 2020 q29
I'm pretty sure these are all on THSC, FYI
Girraween 2020 q3
Manly Selective 2020 q3, 19
Sydney Girls 2020 q7, 36
Killara 2020 q10
Normanhurst Boys 2020 q10, 20
Sydney Boys 2020 q28
Sydney Grammar 2020 q28
North Sydney Girls 2020 q29
North Sydney Boys 2020 q29
I'm pretty sure these are all on THSC, FYI
Doesn't P(A) =1/4 or 5/4
It's a probability, so