An urn contains two red balls and four white balls. Sample successively five times at random and with replacement, so that the trials are independent.

Question

An urn contains two red balls and four white balls. Sample successively five times at random and with replacement, so that the trials are independent.

(a) Compute the probability that no two balls drawn consecutively in the sequence of five balls have the same color.

(b) How would your answer to part (a) change if the sampling is without replacement?

For (a). I am not sure the correct way of thinking this. I was listing all the possible cases of drawing the same ones consecutively and then use complement rule to figure out the probability. Is that the right way of thinking?

Answer 1

Hint: For the first draw, you can start with White or Red. After that, you must draw a ball of the opposite color on every draw. This gives two "valid" sequences: $WRWRW$ or $RWRWR$.

So, the probability for part (a) would be $$\left(\dfrac{4}{6}\right)\left(\dfrac{2}{6}\right)\left(\dfrac{4}{6}\right)\left(\dfrac{2}{6}\right)\left(\dfrac{4}{6}\right) + \left(\dfrac{2}{6}\right)\left(\dfrac{4}{6}\right)\left(\dfrac{2}{6}\right)\left(\dfrac{4}{6}\right)\left(\dfrac{2}{6}\right) = \left(\dfrac{2}{3}\right)^3\left(\dfrac{1}{3}\right)^2+\left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^3 \approx 4.94\%$$ The probability for part (b) would be different because there are not enough red balls for $RWRWR$. This leaves $WRWRW$ as the only possible valid outcome, with probability $$\left(\dfrac{4}{6}\right)\left(\dfrac{2}{5}\right)\left(\dfrac{3}{4}\right)\left(\dfrac{1}{3}\right)\left(\dfrac{2}{2}\right) \approx 6.67\%$$ This may seem surprising because although there are fewer valid outcomes, the probability of its occurrence is actually higher.