Sally's Goal-Shooting - Binomial Distribution Q&A

[psyclone]

Hello all,

I just have a question which covers binomial distribution.

Sally is a goal shooter. Assume each attempt at scoring a goal is independent, in the long term her scoring rate has been shown as 80% (i.e. 80% success rate).

Question:

What's the probability, (correct to 3 decimal places) that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts are successful?

I've solved the problem the following way.

X~ Bi(8,0.8)

Pr(X=6) = 0.8^6*0.2^2 = 0.0105

Pr(X=4) = 0.8^4*0.2^4 = 0.0007


Pr(X=4 | X=6) = Pr(X=4 & X=6) = Pr(X=4) = 0.0007 = 0.067
------------------- ---------- --------- <--------divided signs
Pr(X=6) Pr(X=6) 0.0105

Note:{'&' means Intersection}, the answer given = 0.214. Is that correct?

I would like your thoughts.

Note: The previous two parts of the same question are (where the above is part iii):
(i) chance of getting 8 out of 8 shoots as goals.
(ii) change of getting exactly 6 out of 8 shoots as goals.
ref: MathsQuest(Yr-12)

 

[Simon Bridge]

Need the reasoning.

$p^n(1-p)^{N-n}$ is the probability of n successes out of N trials.
It is not the probability that those n successes are the first n of them.

The probability P(X=6) you gave is the probability that any six are successes right?

If I read you correctly:
The probability P(X=4) you gave is the probability that any four of the trials are successes.

 

[Office_Shredder]

Pr(X=4 | X=6) is zero, because if X=6 it is never equal to 4.

Some food for thought: Does this part of the question even require knowing what Sally's probability of scoring each individual goal is?

 

[haruspex]

$p^n(1-p)^{N-n}$ is the probability of n successes out of N trials.
It is not the probability that those n successes are the first n of them.

Umm.. backwards?
$p^n(1-p)^{N-n}$ is the probability of n successes out of N trials, and those being the first n.
$^NC_np^n(1-p)^{N-n}$ is the probability of n successes out of N trials, those being any n.

 

[Simon Bridge]

@haruspex:
I picked the right issue but got it backwards - thanks.
<annoyed with self>

Advise to OP would be to be more careful about defining events.
Off the wording in post #1,
It looks like X is the number of trials that are successes
So X=x would be the event that exactly x trials were successes.
P(X=x) equation given was for the event that a particular x trials were a success.
P(X=x|X=y) equation... was for different X again?

By the first definition of X: P(X=x|X=y)=0 unless x=y - so that's not what is intended!

How about:
X is the event that the 1st n out of N trials are successes - the remaining N-n trials being anything.
Y is the event that any m out of the N trials are successes
X$\small\cap$Y would be the event that there are m successes in N trials, in which the 1st n trials are successes.
... or something.

 

[psyclone]

sorry for not getting back sooner. Thanks for responses,

To both of the above comments.
Correct. The intersection is equal to 0.

If we consider the problem as a tree diagram, there are 6 branchs of X=6 given the first 4 shots are successfull.
and given;
Pr(x=6) = 0.2936 (exactly 6 isn't the same as only 6, my bad).

Therefore:
(6*0.8^6*.2^2)/0.2936 = 0.214

 

[haruspex]

sorry for not getting back sooner. Thanks for responses,

To both of the above comments.
Correct. The intersection is equal to 0.

If we consider the problem as a tree diagram, there are 6 branchs of X=6 given the first 4 shots are successfull.
and given;
Pr(x=6) = 0.2936 (exactly 6 isn't the same as only 6, my bad).

Therefore:
(6*0.8^6*.2^2)/0.2936 = 0.214

That's the right answer.
The easiest way is to to turn it into: pick 6 objects from 8 in a line; what's the probability you pick the first 4? A: ⁴C₄*⁴C₂/⁸C₆