1
$\begingroup$

Here's the question and my current solution:

Suppose we have two bags each containing black and white balls. Bag A contains 50 black balls, and twice as many white balls as black balls. Bag B contains 30 white balls, and 3 times as many black balls as white balls.

Bag A: 50 black balls, 100 white balls Bag B: 90 black balls, 30 white balls

Suppose we give Bag A to a robot who is programmed to randomly select one ball from the bag, and then put the ball back to the bag after recording its color. After the robot performs the task 3 times, what is the probability that it records 2 black balls and 1 white ball?

MY SOLUTION:

P(black) = 1/3, P(white) = 2/3

Answer = 1/3 * 1/3 * 2/3 = 2/27

Now this is just basic probability, but I'm thinking that I might need to use the Binomial Random variable function, (n choose k) p^k(q^n-k). Did I solve this correctly or is it more complicated that it seems?

Thanks!

$\endgroup$
0

1 Answer 1

1
$\begingroup$

You solved correctly the problem of finding the probability of $2$ black followed by $1$ white (BBW). If by $2$ black and $1$ white you mean any order is OK, then we need to add in the probabilities of BWB and WBB. Each of these is the same as the probability of BBW, so we get $3$ times your answer.

It is this kind of analysis that leads to the binomial coefficient $\binom{n}{k}$ in the formula you quoted. That takes into account the fact that $k$ successes and $n-k$ failures can happen in $\binom{n}{k}$ orders.

If the probability of success on any one trial is $p$, and the probability of failure is $q=1-p$, and the trials are independent, then the probability of any particular string of $k$ successes and $n-k$ failures is $p^kq^{n-k}$.

But there are $\binom{n}{k}$ such strings, which yields the probability $\binom{n}{k}p^kq^{n-k}$.

$\endgroup$
4
  • $\begingroup$ OK, so in this case it would be (150 choose 3) (4/27)^3 (23/27)^147? $\endgroup$ Feb 25, 2014 at 2:05
  • $\begingroup$ No, you had it almost right before. The robot picked $3$ times, so the number $n$ of trials is $3$. We want the probability of $2$ black (and therefore $1$ white). The probability of black is always $\frac{50}{150}$, that is, $1/3$, so $p=1/3$. The probability is $\binom{3}{2}(1/3)^2(2/3)^1$. Just $3$ times the answer you found, as mentioned in the answer. $\endgroup$ Feb 25, 2014 at 2:13
  • $\begingroup$ Thank you Andre! Very helpful! $\endgroup$ Feb 25, 2014 at 2:18
  • $\begingroup$ You are welcome. $\endgroup$ Feb 25, 2014 at 2:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .