Probabilities

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Tamara33

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Oct 30, 2010
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Hello everyone!

I'm working on a problem that requires me to calculate a probability.

Specifically, if I had eight envelopes (only one of which contains a $100), what is the probability that I will pull the envelope with the bill ON the 5th try.

I used a tree diagram and ended up multiplying 7/8 * 6/7 * 5/6 * 4/5 * 1/4 - which gives me 1/8! I don't think this is correct since this would be the probability of selecting it on the first try. Any help would be greatly appreciated!

I hope you're all planning a great weekend.
 
Hi

The answer is right because it should not matter how many tries you make: overall
the probability will always be the same, 1/8, always. You should not be "luckier" if
you only had, say 3 tries.
Formally if N is the total number of items and you select without replacement k items,
the overall probability is

(N-1) (N-2) (N-k+1) 1
------ ------- .... -------- ........... = (1/N)
N (N-1) (N-k+2) (N-k+1)
 
With or without replacement does not matter in this case because there is only one envelope with the money in it, the outcome is binary, and all the envelopes are the same.
 
qso said:
With or without replacement does not matter in this case because there is only one envelope with the money in it, the outcome is binary, and all the envelopes are the same.
Click to expand...

As I see it - it should matter.

With replacement (meaning the pulled envelope is back in the stack) ? the probability is: (7/8)^4 * (1/8)
 
Of course it matters.

If the envelope is put back the trials are independent.

As SK showed, one can use a geometric distribution to find how many failures occur before a single success.

If x is the # of failures before a success, then

\(\displaystyle P(x)=(1-p)^{x}\cdot p\)

\(\displaystyle P(4)=(\frac{7}{8})^{4}\cdot \frac{1}{8}\approx .073\)

The geometric distribution is a special case of the negative binomial distribution, which is used for probabilities such as this.
 
Think about it.
If you replace the envelope you start over - each trial is independent. The prob. is 1/8.
 
Howevwer, you must fail 4 times and must suceed the fifth time - those are a priori conditions.

We are not discussing chance of success at "any" time.
 
You are answering a different question to that which is posed.
Envelopes don't have memories.
The question asks what is the probability of success ON the 5th try.
You could draw an envelope a million times but at each trial there will
always be 8 envelopes and 1 with the money if you are doing it with replacement.

The question you are answering is that if you make 5 withdrawls with replacement
what is the probability that 4 of them will fail and 1 will succeed. That is not the
question posed. If is then the question needs to be better worded. But the word "ON"
was capitalized in the question.
 
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