Probability - When Math meets English!

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lingua

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Hello,

thanks for taking your time to check this problem I have with a conditional probability question.

Qn. Of 50 people surveyed, 35 played tennis and 26 played netball. Everyone surveyed played at least one of these sports.
If one person is selected at random, find the probability that he/she plays tennis given that he/she also plays netball.

I need to use the formula Pr(A|B) = Pr(A and B)/Pr(B).

I know the answer (if the back of the book is correct) is 11/35.

Please help! :wink:

Thankyou :p
 
Based upon the given, the ‘back of’ the book cannot be correct!
P(T)=35/50, P(N)=26/50, and P(T∪N)=1.
P(T∪N)=P(T)+P(N)-P(T∩N) so P(T∩N)=11/50.

P(T|N)= P(T∩N)/P(N).
 
Sorry, I can't view all of you characters used in your workings,
this is the answer I came to:
Let T = Probability of a person playing tennis
Let N = Probability of a person playing netball
Pr(T|N) = Pr(T∩N)/Pr(N)
Pr(T) = 35/50, Pr(N) = 26/50
35Tennis+26Netball = 61 , therefore 11 people do both T and N.
Pr(T∩N) = 11/50

Therefore, Pr(T|N) = (11/50)/(26/50) = 11/26

- I have a feeling BOB is wrong, could you please tell me if all of my workings are correct then, I suppose that is the main thing.

Thankyou! :p
 
Re: Conditional Probability - URGENT

Hello, lingua!

Your work is correct!

A Venn diagram will make it even clearer.

Of 50 people surveyed, 35 played tennis and 26 played netball.
Everyone surveyed played at least one of these sports.
If one person is selected at random,
find the probability that he/she plays tennis given that he/she also plays netball.
Click to expand...
Code: 
      -----  -----
     /  T  \/  N  \
    /      /\      \         All 50 people are accounted for
   /      /  \      \             in the Venn diagram.
  |      |    |      |
  |  24  | 11 | 15   |            24 play only tennis,
  |      |    |      |            15 play only netball,
   \      \  /      /             11 play both sports.
    \      \/      /
     \     /\     /
      -----  -----
Given that the chosen person plays netball,
. . the choices are reduced to the 26 who play netball.

Of those 26 people, there are 11 who also play tennis.

Therefore: . P(tennis | netball) is undeniably 11/26


I'd say that the BoB has an egregious error.
 
Oh! Great!
Thanks heaps for your help!
I've got a Math exam today (that's why the question & my understanding was urgent), and I feel alot more confident!
Thanks again!

8-)
 
Just one more question... :p

One letter is randomly selected from each of the words HOORAY FOR MATHS. Find the probability of getting at least one vowel which is not the letter O.
- I hate asking about probability, it's seems so easy (and the maths is), but the concepts or what the question is asking of you is sometimes very hard to grasp.
*P.S. Sorry if you're American, I know you say "math" rather than maths... Just add 1 more letter! hehehe.. :lol: (I'm Aussie by the way!)

Thanks again... :wink:
 
lingua said:
Just one more question... :p
One letter is randomly selected from each of the words HOORAY FOR MATHS. Find the probability of getting at least one vowel which is not the letter O.
Click to expand...

Well, only the "a" in hooray and in "maths":
1/6 from "hooray" times 1/5 from "maths" = 1/30
 
Thanks Denis,
although my trustworthy (NOT!) book says 1/3. Arrggh... Don't they check these mistakes when the make new editition (or maybe before they publish the first? hahaha. . . :eek: )
But, if I'm not mistaken, that's the probability of getting a vowel (not O), twice, but it says at lease one vowel so I think we may need to factor this in somehow. . . any ideas? :?
 
Not my best subject but I would say:
The probability of NOT getting the A from hooray is 5/6
The probability of NOT getting the A from maths is 4/5
The probability of NOT getting either A is (4/5)(5/6) = 4/6 = 2/3
so 1 - 2/3 = 1/3 is what you are looking for.
Apologize to the editor :twisted:
 
hahahaha. . . .
I refuse to appologise, check out my question in previous posts under this subject, it was from the same author.
But, thanks alot for your help!

P.S. Why can't I do it the other way around like denis, and come up with the same answer?
 
Hello, lingua!

Why can't I do it the other way around like denis, and come up with the same answer?
Click to expand...
Denis found the probability of getting both A's.

Gene had the best idea.
Whenever a problem says "at least", consider the opposite ("none").
Often the task is much easier.

Here's my version of his approach.

There are 6 choices of letters from HOORAY.
There are 3 choices of letters from FOR.
There are 5 choices of letteres from MATHS.
. . . Hence, there are 6 x 3 x 5 = 90 possible selections that could be made.

Of these 90 selections, how many have <u>no</u> A's?
From HOORAY, there are 5 choices.
From FOR, there are 3 choices.
From MATHS, there 4 choices.
. . . So, there are 5 x 3 x 4 = 60 ways to get no A's.

Hence, there are 90 - 60 = 30 ways to get at least one A.

Therefore: .P(at least one A) .= .30/90 .= .1/3
 
Okay, I'm flexable.
The probability of A from h(ooray) and not from m(aths) is 1/6*4/5
The probability of A from m and not from h is 5/6*1/5
The probability of A from both is 1/6*1/5
Add them up.
Since there are only three it is feasible. In a longer sentence...
 
let x= number that only played only tennis
let y = number that played both tennis and netball
let z= number that played only netball

eq1 x+y=35
eq 2 y+z=26
eq3 x+y+z=50 from eq1 x=35-y from eq2 z= 26-y substitute

[35-y]+y+[26-y]=50 clear brackets
61 -y=50
y=11

prob that a person played both 11/50
prob that a person played tennis 35/50
prob that person chosen played tennis, and netball:
11/50 divided by 35/50
11/35 answer

redo it using the correct notation now that you see the method
Arthur
 
Ya...I somehow read that as "go get the 2 A's"!

Should be:

I pick A from hoorAy: 1/6

5/6 of time I won't; so I go pick from mAths: 5/6 * 1/5 = 1/6

1/6 + 1/6 = 2/6 = 1/3
 
the answer is 1/3
the probability of not getting a vowel from 1st word = 5/6
the probability of not getting a vowel from 2nd word=1
the probability of not getting a vowel from the 3rd word =4/5

the probability of not getting a vowel from all 3 words = 5/6*1*4/5 = 2/3
the probability of a vowel =1-2/3
1/3 answer
Arthur
 
Great,
that was the answer I was looking for Arthur, nice and simple.
Thanks!
:lol:
 
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