Conditional probability (1 Viewer)

Aerlinn

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I'm just a little confused about this topic. I understand the thinking process when working out these problems, or by drawing tree diagrams and such, just dont know how to use the formula P(A given B)= P(A intersecting with B)/ P(B).
Working out P(A intersecting with B) is a problem because multiplying A * B only applies for independent events. So when I come across harder problems, I get the answer wrong.
As an example, the probability that a full forward in Australian Rules football will kick a goal from outside the 50m line is 0.15. If the full forward has 10 kicks at goal from outside the 50metre line, find the probability he will kick more than one goal, given that he kicked at least one goal. (ans: 0.5674)
Help= :) :wave:
 

n.gallagher

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What the hell??? Luv, I’ve never seen a question like that before, and I’ve done every single HSC exam and every trial paper I can get my hands on…also, I have never come across that equation except when I was looking through a 2 unit math text book (it’s not in the General math syllabus)…either your in the wrong forum, or your teacher is telling you crap…but, just out of curiosity, where did you get that question from?
 

PC

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Aerlinn said:
As an example, the probability that a full forward in Australian Rules football will kick a goal from outside the 50m line is 0.15. If the full forward has 10 kicks at goal from outside the 50metre line, find the probability he will kick more than one goal, given that he kicked at least one goal. (ans: 0.5674)
Help= :) :wave:
The probability that the player will kick any goal from outside the 50 metre line is 0.15.
Therefore the probability that he will miss a goal from outside the 50 metre line is 1 – 0.15 = 0.85.

Now the probability that he will miss 10 goals in a row = 0.85 x 0.85 x 0.85 x ... x 0.85 = (0.85)^10 = 0.1968744043

The probability that he will miss 9 goals and kick one = 10 x (0.85)^9 x 0.15 = 0.3474254194.
The 10 comes from the 10 ways that he can kick one goal, 1st, 2nd, ..., 10th attempt.

So ... the probability that he will kick at least two goals is basically everything except missing all of them or kicking just one.
So probability = 1 - 0.1968744043 - 0.3474254194
= 0.45557001762
= 0.4556

Not sure where your answer comes from.

It should be noted that questions like these are well outside the scope of the General Mathematics course and are more suited to the 3 unit topic of the Binomial Theorem.
 
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Captain Gh3y

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Try again PC

What you found is the answer to the question "What is the probability that he will kick at least two goals"

when the question was "What is the probability that more than one goal will be kicked, given that at least one goal is kicked".

The probability (A given B), is equal to the probability of (A and B) divided by the probability of B.
or P(A given B) = P (B and A) / P(B).

B is "at least one goal is kicked" and A is "at least two goals are kicked". Note that in this example, A is a subset of B, so P(B intersecting with A) is equal to P(A).

First, the probability that he will kick at least one goal is
B = 1 - 0.85^10
B = 0.8031255957

Second, the probability that at least two goals are kicked is

P(B and A) = 1 - 0.85^10 - (10C1 * 0.85^9 * 0.15^1)
P(B and A) = 0.4557001762
This is what PC (incorrectly) calculated.


Then applying the formula, P(A given B) = 0.4557/0.80312

P(A given B) = 0.5674 as expected.
 
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n.gallagher

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PC said:
It should be noted that questions like these are well outside the scope of the General Mathematics course and are more suited to the 3 unit topic of the Binomial Theorem.
Even so, still realise that this question is way outside the scope of the general math syllabus, and your teacher should not be teaching you this at all. Considering that your teacher is gay, don’t listen to him/her/shim and only refer to the syllabus, that way you won’t be wasting your time learning 3 unit calculations.
 

PC

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Correction noted.

Aye aye Captain!

Discrete Maths. Are you at Newcastle Uni? I did that topic!
 

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