Probability of 3rd ball being green while drawn without replacement, in a case where there are 5 blue 3 green and 7 yellow marbles in an urn. I'm having a hard time coming to a conclusion about the process of the math. I drew a probability tree to have a better understanding, the final branch being P(green) and P(green)', then I added the probabilities of the 3rd one being green and divided it by all possible outcomes of the third pick. My answer comes to 7/39 but I'm not sure if my process is correct.
Probability – Probability of Drawing a Green Marble from an Urn
bayes-theoremconditional probabilityprobability
Related Solutions
$1.$ With replacement: There are three ways this can happen, red, red, red; blue, blue, blue; and green, green, green.
The probability of red, red, red is $\left(\dfrac{5}{19}\right)^3$. Find similar expressions for the other two colurs, and add up.
$2.$ Without Replacement: The probability of red, red, red is $\left(\dfrac{5}{19}\right)\left(\dfrac{4}{18}\right)\left(\dfrac{3}{17}\right)$. Find similar expressions for the other two colours, and add up.
Remark: We could also do the problem by a counting argument. Let's look at the first problem. Put labels on the balls to make them distinct. Then for with replacement, there are $19^3$ strings of length $3$ of our objects. (The three objects in the string are not necessarily distinct.) All of these $19^3$ strings are equally likely.
There are $5^3$ all red strings, $11^3$ all blue, and $8^3$ all green. This gives probability $\dfrac{5^3+6^3+8^3}{19^3}$.
One can do a similar calculation for the without replacement case.
For without replacement, there is a third approach. We can choose $3$ objects from $19$ in $\dbinom{19}{3}$ ways. And we can choose $3$ of the same colour in $\dbinom{5}{3}+\dbinom{6}{3}+\dbinom{8}{3}$ ways.
Your approach is correct, since you are covering all possible events that satisfy the condition, and you are not over or undercounting.
An alternate perspective, which can confirm your answer, is that we have $ 6 \times 5=30 $ possible ways of conducting this draw. Of this, to satisfy our condition, we need the first ball to be white and the second one to be green, or vice versa. There are $ 2 \times 1$ ways to do both of the above, so 4 ways total.
We just enumerated all the possible cases, and all the possible cases that satisfy your probability, and we obtain 4/30. Note that I am summing 2+2 and then dividing by 30 in the same way that your method considers two equal probabilities. It is pure coincidence that they are equal.
Best Answer
Probability of case I $= \frac{12}{15}\cdot \frac{11}{14} \cdot \frac{3}{13}=\frac{66}{455}$
Probability of case II $= \frac{12}{15}\cdot \frac{3}{14} \cdot \frac{2}{13}=\frac{12}{455}$
Probability of case III $= \frac{3}{15}\cdot \frac{2}{14} \cdot \frac{1}{13}=\frac{1}{455}$
Probability of case IV $= \frac{3}{15}\cdot \frac{12}{14} \cdot \frac{2}{13}=\frac{12}{455}$
Total Probability$=\frac{66}{455}+\frac{12}{455}+\frac{1}{455}+\frac{12}{455}=\frac{91}{455}=\boxed{\frac{1}{5}}$