there are 300 candies in a jar: 150 red, 100 green, 50 yellow. 10 candies are taken out, randomly, one by one without returning them to the jar.
I need to calculate the probability that:
- The first yellow candy in taken out at the 5th pick
- if the first yellow was taken out 5th, then the 6th pick is also yellow
- if out of the 10 candies exactly 2 are yellow, then exactly 4 candies are red
- if the first and second are red, then exactly 6 are red
(1) let $Y_i$ be the event that the first yellow was chosen at the pick no. i, so the probability that the first 4 candies weren't yellow is $$P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c) = (1-\frac{50}{300})(1-\frac{50}{299})(1-\frac{50}{298})(1-\frac{50}{297})\approx 0.48$$ and
$$P\{first\space yellow\space was\space 5th\}=P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5)=P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c)P(Y_5) = 0.48 * \frac{50}{296} \approx 0.08$$
(2) let E be the event that yellow was picked 6th. $$P\{6th\space pick\space is\space yellow|first\space yellow\space was\space 5th\}=\frac{P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5\cap E)}{P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5)}=\frac{P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5)\cdot \frac{49}{295}}{P(Y_1^c\cap Y_2^c\cap Y_3^c\cap Y_4^c\cap Y_5)} = \frac{49}{295} \approx 0.166 $$
but I'm struggling with (3) and (4).
for (3) I know I can look at $$P\{exactly\space 4\space reds | exactly\space 2\space yellows\} = \frac{P(\{exactly\space 4\space reds\}\cap \{exactly\space 2\space yellows\})}{P\{exactly\space 2\space yellows\}}=\frac{P\{exaclt\space 4\space greens\}}{P\{exactly\space 2\space yellows\}}$$ but I don't know how I can calculate those probabilities without a ton of manual labor as the probability for every pick depends on the order of each of the picks and the color of the candies picked. Same for (4), calculating the probability of choosing exactly 4 reds out of the remaining 8 became super complex. would falling back to "basic" combinatorics simplify this?
Best Answer
Your answer is correct. Here is another method. If the first yellow candy is selected fifth, then the first four candies must be selected from the $300 - 50 = 250$ candies which are not yellow, then a yellow must be selected from the $296$ candies which remain. $$\Pr(\text{fifth pick is first yellow}) = \frac{\dbinom{250}{4}}{\dbinom{300}{4}} \cdot \frac{50}{296}$$
Your answer is again correct.
For exactly two of the first ten candies to be yellow, eight of the other $250$ candies must be selected. For exactly two of the first ten candies to be yellow and exactly four of the first ten candies to be red, the other four candies must be green. If we let $R$ be the event that exactly four of the first ten candies are red and $Y$ be the even that exactly two of the first ten candies are yellow, then $$\Pr(R \mid Y) = \frac{\dbinom{150}{4}\dbinom{100}{4}\dbinom{50}{2}}{\dbinom{50}{2}\dbinom{250}{8}}$$
For this to occur, we must first pick two red candies and then select four red candies among the next eight candies. If the first two candies are red, we are left with $148$ red candies, $100$ green candies, and $50$ yellow candies. Let $R$ be the event that the first two candies are red; let $E$ is the event that the first two candies are red and exactly six of the first ten candies are red. Given the first two candies are red, event $R$ occurs if we first pick any eight of the remaining $148 + 100 + 50 = 298$ candies and event $E$ occurs if we select exactly four of the remaining $148$ red candies and exactly four of the other $100 + 50 = 150$ candies with the remaining eight selections. $$\Pr(E \mid R) = \frac{\dbinom{148}{4}\dbinom{150}{4}}{\dbinom{298}{8}} $$