# Related Independent Events (2.5.24)

Please verify my solution to this problem.

Each of m urns contains three red chips and four white chips. A total of r samples with replacement are taken from each urn. What is the probability that at least one red chip is drawn from at least one urn?

Here is my solution:

$$P$$(at least one red chip is drawn from at least one urn)$$=$$
$$1-P$$(no red chip is drawn from any urn)$$=$$
$$1-P$$(only white chips are drawn from the urns)$$=$$
There is a $$\tfrac47$$ chance of drawing a white chip out of any one of the m urns.
Since it is replaced, there is an $$(\frac47)^{r}$$ chance of drawing a white chip for each of the r draws from any urn.
Therefore, $$P$$(draw at least one red chip from at least one urn) $$=$$ $$1-(\frac47)^{m*r}$$

• Where's your solution? – Shubham Johri Aug 2 at 4:17
• It is very difficult to verify invisible solutions – Graham Kemp Aug 2 at 4:24
• I'm new to the site. Now I realize that my solution is included in the question, not as an answer. – Dan Banaszak Aug 2 at 4:54

The complement event is we do not see any red chips, which means that we only get white chips in all of the $$m\cdot r$$ trials.
Hence, the answer is $$1-\left( \frac47\right)^{m\cdot r}$$