I am trying to solve this problem.
There are four identical red boxes and three identical blue boxes. If the boxes are randomly selected one at a time without replacement, find the probability that number of boxes which must be selected in order to obtain all three blue boxes is at least five.
I tried
- We select 3 boxes and all boxes are blue. In this case, we have one way.
- We select 4 boxes and there are three blue boxes, then we have three cases: RED BLUE BLUE BLUE, BLUE BLUE RED BLUE, BLUE RED BLUE BLUE.
- We select 5 boxes and there are three blue boxes. The end box must blue. The number of select is
C(4, 2) = 6. - We select 6 boxes and there are three blue boxes. The end box must blue. The number of select is
C(5, 3) = 10. - We select 7 boxes and there are three blue boxes. The end box must blue. The number of select is
C(6, 4) = 15.
The probability to select 3 boxes such that the number of boxes is greater than or equal to 5 is
$$(6 + 10 + 15)/(1 + 3 + 6 + 10 + 15)=31/35.$$
Is this correct?
If the blue boxes and red boxed are different. How to find The probability to select 3 boxes such that the number of boxes is greater than or equal to 5?
Best Answer
New Interpretation
There is disagreement about what the OP is asking. It has been suggested that my original interpretation is wrong. In hindsight, it does seem very plausible that I was mistaken. Therefore, I left my original interpretation at the very end of my answer, just in case it turns out to be what the OP intended.
The new interpretation is to compute the probability that in the first $4$ boxes selected (without replacement), less than $3$ blue boxes were selected. This would correspond to there needing to be at least $5$ boxes selected, before all of the blue boxes were selected.
The number of ways that $4$ boxes can be selected out of $7$ is $\displaystyle \binom{7}{4} = 35$.
The number of ways that $m$ blue boxes were among the $4$ selected boxes [with $m \in \{0,1,2,3\}$] is
$$\binom{3}{m} \times \binom{4}{4 - m}.$$
This is based on the idea that if there were $m$ blue boxes (out of $3$ total blue boxes) selected, then there were $(4 - m)$ red boxes (out of $4$ total red boxes) selected.
Therefore, the desired computation is
$$\frac{1}{\binom{7}{4}} \times $$
$$\left[\binom{3}{0} \times \binom{4}{4}\right] ~+~ \left[\binom{3}{1} \times \binom{4}{3}\right] ~+~ \left[\binom{3}{2} \times \binom{4}{2}\right].$$
This equals $$\frac{1}{35} \times [1 + 12 + 18] = \frac{31}{35}.$$
Edit
A shortcut, based on my approach would be to compute $~1 - ~$ the probability that in $4$ selected boxes, exactly $3$ of them were blue.
This would be
$$1 - \frac{\binom{3}{3} \times \binom{4}{1}}{\binom{7}{4}} ~=~ 1 - \frac{4}{35} = \frac{31}{35}.$$
Old Interpretation
I am interpreting the OP (i.e. original poster's) intent to be that you are given that at least $5$ boxes were selected (without replacement), and wish to calculate the probability that all of the $3$ blue boxes were selected.
Let $k$ denote the number of boxes that were selected, where $k \in \{5,6,7\}$.
Based on this, let $p(k)$ denote the (relative) probability that $k$ boxes were selected.
Assuming that $k$ boxes were selected, let $f(k)$ denote the probability that all $3$ blue boxes were selected.
Then, the desired computation is
$$\sum_{k=5}^7 \left[p(k) \times f(k)\right].\tag1 $$
There are clearly $2^7$ equally likely possibilities of the (raw) selection of boxes, because (presumably) each box either was selected [probability $= (1/2)$] or was not selected [probability $= (1/2)$].
Therefore, the relative probabilites are $[\binom{7}{7} = 1]$ versus $[\binom{7}{6} = 7]$ versus $[\binom{7}{5} = 21]$ with respect to the relative probabilities of either $7, 6$, or $5$ boxes being selected.
Therefore,
$$p(7), p(6), p(5) ~=~ \frac{\binom{7}{7}}{29}, \frac{\binom{7}{6}}{29}, \frac{\binom{7}{5}}{29} ~~\text{respectively}.\tag2 $$
If $k$ boxes were selected [with $k \in \{5,6,7\}$], then the chance that all $3$ blue boxes were selected is
$$f(k) = \frac{\binom{4}{k-3}}{\binom{7}{k}}. \tag3 $$
The explanation for this is that in (3) above, the denominator represents the number of ways that the $k$ boxes could be selected. Further, if all $3$ blue boxes were selected, then $k-3$ red boxes were selected from the 4 red boxes. Therefore, the numerator reflects the number of ways of choosing $k-3$ red boxes from the $4$ red boxes.
Utilizing the expressions in (1), (2), and (3) above, the desired computation is
$$\sum_{k=5}^7 \left[p(k) \times f(k)\right]$$
$$ ~=~ \left[\frac{\binom{7}{7}}{29} \times \frac{\binom{4}{4}}{\binom{7}{7}}\right] ~+~ \left[\frac{\binom{7}{6}}{29} \times \frac{\binom{4}{3}}{\binom{7}{6}}\right] ~+~ \left[\frac{\binom{7}{5}}{29} \times \frac{\binom{4}{2}}{\binom{7}{5}}\right].\tag4 $$
Simplifying the math in (4) above gives
$$\frac{1}{29} \times [1 + 4 + 6] = \frac{11}{29}.$$