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Help anyone? Probability

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been stuck on this a while ;/
Screen Shot 2017-10-23 at 17.03.56.png
Original post by marinacalder
Attachment not found
been stuck on this a while ;/
Original post by marinacalder
...


You have the following probabilities, letting P(i)P(i) where i∈{m,w,c,t}i \in \{m,w,c,t \} being the probability of choosing either a man, woman, someone who plays cricket, or someone who plays tennis.

P(w)=0.6P(w)=0.6 and P(m)=0.4P(m)=0.4

P(m∣c)=25P(m | c) = \frac{2}{5}

P(c∣w)=23P(c | w) = \frac{2}{3}

You want to find P(w∣t)P(w | t).

Any thoughts/attempts?
Original post by RDKGames
You have the following probabilities, letting P(i)P(i) where i∈{m,w,c,t}i \in \{m,w,c,t \} being the probability of choosing either a man, woman, someone who plays cricket, or someone who plays tennis.

P(w)=0.6P(w)=0.6 and P(m)=0.4P(m)=0.4

P(m∣c)=25P(m | c) = \frac{2}{5}

P(c∣w)=23P(c | w) = \frac{2}{3}

You want to find P(w∣t)P(w | t).

Any thoughts/attempts?

wait, so not P(woman intersect tennis)?
Original post by marinacalder
wait, so not P(woman intersect tennis)?


No because it's conditional probability.

You have P(w∣t)=P(w∩t)P(t)\displaystyle P(w |t )=\frac{P(w \cap t)}{P(t)} and since P(t)≠1P(t) \neq 1 you do not have P(w∣t)=P(w∩t)P(w |t)=P(w \cap t). If the only sport was tennis, then yes you would have what you got there.
(edited 6 years ago)
Original post by RDKGames
No because it's conditional probability.

You have P(w∣t)=P(w∩t)P(t)\displaystyle P(w |t )=\frac{P(w \cap t)}{P(t)} and since P(t)≠1P(t) \neq 1 you do not have P(w∣t)=P(w∩t)P(w |t)=P(w \cap t). If the only sport was tennis, then yes you would have what you got there.



oh gosh, is there a way to really quickly work out whether it's conditional or not? I have trouble with this..
Original post by marinacalder
oh gosh, is there a way to really quickly work out whether it's conditional or not? I have trouble with this..


Yeah you just do it from the context... conditional is like saying 'OK here are two things (humans and sports, in this ex) that do not influence each other's probabilities of being chosen (ie if I pick a woman, I can still pick any sport I want that is available, whereas I cannot then pick a man)'

EDIT: Also the word 'given' is a trigger word for conditional probability in a question. i.e. 'What is the prob. of choosing event A GIVEN event B' - though this doesn't seem to be the case here. Then again, probability isn't really my thing.
(edited 6 years ago)
Original post by RDKGames
No because it's conditional probability.

You have P(w∣t)=P(w∩t)P(t)\displaystyle P(w |t )=\frac{P(w \cap t)}{P(t)} and since P(t)≠1P(t) \neq 1 you do not have P(w∣t)=P(w∩t)P(w |t)=P(w \cap t). If the only sport was tennis, then yes you would have what you got there.


Also, if it's conditional, does that mean that trying to work it out assuming there are 100 total members won't work? and if so , why is that?
Original post by marinacalder
Also, if it's conditional, does that mean that trying to work it out assuming there are 100 total members won't work? and if so , why is that?


What's the point in assuming the sample size? Probabilities here aren't influenced by this.
Original post by RDKGames
What's the point in assuming the sample size? Probabilities here aren't influenced by this.


Would you mind providing a solution to this?:smile:
2/5 are men who play cricket so 2/5 of 40% = 16%2/3 of the cricketing members are women - ie twice as many as the men , so 32%remaining men play tennis 3/5 of 40% =: 24%Which leaves 100- 16- 32- 24 = 28% women who play tennis ( or 7/25 )
Original post by marinacalder
Would you mind providing a solution to this?:smile:


I misinterpreted the info when it came to the probabilities I gave on conditionals, they should be P(c∣m)=25P(c |m)=\frac{2}{5} and P(w∣c)=23P(w | c) = \frac{2}{3}, and yes the question asks for P(w∩t)P(w \cap t) after looking at it properly - sorry, been doing too many conditionals lately!

Then you know that P(c∩m)=P(c∣m)P(m)=25⋅0.4=0.16P(c \cap m )=P(c|m)P(m)=\frac{2}{5}\cdot 0.4=0.16

Then you know that P(c∩w)=P(w∣c)P(c)=23P(c)P(c \cap w)=P(w|c)P(c)=\frac{2}{3}P(c)

So then you have P(c∩m)+P(c∩w)=P(c)⇒0.16+23P(c)=P(c)P(c \cap m)+P(c \cap w)=P(c) \Rightarrow 0.16+\frac{2}{3}P(c)=P(c) which you can solve for P(c)P(c).

Then similarly get the probability for P(w∩t)=P(w)−P(w∩c)P(w \cap t)=P(w)-P(w \cap c)

Drawing up a table and doing this might be more intuitive.
(edited 6 years ago)

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