3
$\begingroup$

I would appreciate help with the following problem, since I can't quite figure out the effect an increasing number of trials has on probability:

Suppose a bin has white marbles and black marbles. Say the probability of choosing a black marble is $P(B) = \beta$. Each experiment consists of taking 5 marbles from this bin. Certainly, the probability that we get no black marbles from one experiment is $(1-\beta)^5$.

Question: If we repeat this experiment of 5 marbles at a time, with replacement, $N$ times then what is the probability that at least one of our $N$ experiments consists of no black marbles (i.e. at least one of our selections is exactly 5 white marbles)? Also, how does this probability grow with $N$?

References, e.g. books or online notes, addressing this theme would also be appreciated!

$\endgroup$
1
  • $\begingroup$ Welcome at math.SE! It might make things easier if you summarize your five-picks-experiment as one process with sucess probability $\beta'=(1-\beta)^5$. $\endgroup$
    – flonk
    Mar 6, 2014 at 13:51

2 Answers 2

0
$\begingroup$

Let us calculate the probability that none of our selections contain $5$ white marbles. This probability is equal to $\bigl(1-(1-\beta)^5\bigr)^N$ since events are independent. The probability that at least one of our selections is exactly $5$ white marbles is equal to $$ 1-\bigl(1-(1-\beta)^5\bigr)^N. $$

$\endgroup$
3
  • $\begingroup$ thank you! with $\alpha = P(W)$, probability of white, so $\alpha+\beta= 1$, then $P$(no selection of $N$ experiments has exactly 5 white) $=1-(1-\alpha^5)^N$. This leads me to your final answer. thanks again $\endgroup$
    – fractalEd
    Mar 6, 2014 at 13:58
  • $\begingroup$ does one say the "growth" of $1-(1-\alpha^m)^N$, for $m$ fixed, is "exponential in $N$"? $\endgroup$
    – fractalEd
    Mar 6, 2014 at 14:00
  • $\begingroup$ @fractalEd You're welcome! I would say that it converges to $1$ exponentially fast if $0<\beta<1$. $\endgroup$
    – Cm7F7Bb
    Mar 6, 2014 at 14:17
0
$\begingroup$

Increasing the number of trials will impact probability if you are sampling without replacement. This makes intuitive sense. If I do not replace after I draw a sample, the probability increases since the fraction of sample over population increases (population went down). If I sampled with replacement the population would stay the same, thus the probability would remain unaffected.

For example, if my name is in a hat with 6 other names, I have a 1/7 chance of drawing my name. If I replace my name into the hat, then pass the hat to you, you also have a 1/7 chance of drawing your name. No increase in probability. However, should I not replace my name after drawing it from the hat, then pass the hat to you, you would now have a 1/6 chance of drawing your name. Sampling without replacement increases the probably.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.