An urn contains five balls labeled 1 through $5.$ Three balls are removed from the urn one by one randomly, and their numbers recorded in order. No balls are put back in the urn. Assume that all outcomes of the experiment are equally likely.
(i) Let $A$ be the event that the second ball drawn is ball $4$ or $5.$ Find $P(A).$
(ii) Let $B$ be the event that the sample of three balls contains ball $1$ or ball $2$ or both. Find $P(B).$
(iii) Let $C$ be the event that the three recorded numbers come in increasing order. Find $P(C).$
For part (i) I have that computed the probability to be $P(A)= \large\frac{4}{5}\cdot\frac{1}{4}+\frac{4}{5}\cdot\frac{1}{5}.$ I do this because we can decompose the event in question into the event in which ball $4$ is drawn in the second trial, and the event in which ball $5$ is drawn in the second trial, which are disjoint.
For part (ii) I have that we can decompose the event into three disjoint events, which are the simpler events listed. It follows that the probability of the event in question is equal to $2\large \left( \frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\right)+\frac{2}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}.$
For part (iii) I have that there are $11$ possible was to choose the numbers in increasing order. Then we have that the probability is $11$ divided by the total number of choosing $3$ balls out of $5$ without replacement.
Is the above reasoning correct? If not, could you please point out my mistakes?
Best Answer
Since there is no information on the first draw, this probability simply equals: $$P(A) = \frac{2}{5}$$ Alternatively, you could distinguish between drawing a 4 or 5 in the first turn or not. We can then calculate $P(A)$ as: $$P(A) = \frac{2}{5} \frac{1}{4} + \frac{3}{5} \frac{2}{4} = \frac{8}{20} = \frac{2}{5}$$
The probability of drawing 1 or 2 or both is 1 minus the probability of drawing none: $$P(B) = 1 - \frac{3}{5} \frac{2}{4} \frac{1}{3} = 1 - \frac{1}{10} = \frac{9}{10}$$
There are ${5 \choose 3} = 10$ ways to choose the three numbers, and only one correct way to draw them. We thus find: $$P(C) = \frac{{5 \choose 3}}{5 \cdot 4 \cdot 3} = \frac{10}{60} = \frac{1}{6}$$