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. Compute the probability of each of the two sequences

WWRWR

RWWWR

Please explain. I have the correct answer but i do not understand how to get it.

Answer 1

Hint :

If $A$ and $B$ two independant events, $P(A \cap B) = P(A)*P(B)$

Answer 2

The probability to draw a white ball is $\frac{4}{6}=\frac{2}{3}$.

The probability to draw a red ball is $\frac{2}{6}=\frac{1}{3}$.

After drawing there is replacement so that these probabilities stand each of the $5$ times.

Then the chance to draw e.g. RWRRR equals $$\frac{1}{3}\times\frac{2}{3}\times\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}$$ Can you take it from here?

Answer 3

$P(WWRWR) = P(W)P(W)P(R)P(W)P(R) = \frac{4}{6} \frac{4}{6} \frac{2}{6} \frac{4}{6} \frac{2}{6}$

$P(WWRWR) = P(W)P(W)P(R)P(W)P(R)$ follows by independence.

where W = event a white ball is drawn, R = event a red ball is drawn

WWRWR = event a white ball is drawn the fisrt time, a white ball is drawn the second time, a red ball is drawn the third time, a white ball is drawn the fourth time and a red ball is drawn the fifth time