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A green flower will blossom at some point uniformly at random in the next 10 days and be in bloom for exactly 4 days. Independent of the green flower, the red flower will blossom at some point uniformly at random in the next 10 days and be in bloom for exactly 2 days. Compute the probability that both flowers will simultaneously be in bloom at some point in time.

My approach was to fix the day the red flower blossoms from day 1 to 10 and, for each day, determine what days the green flower needs to blossom for it to overlap with the red's bloom. For example, if the red flower blossoms on day 1, the green flower must blossom on either day 1 or 2, a 2/10 chance. I then summed up all the conditional probabilities multiplied by 1/10 to get 43/100 as my answer.

I haven't seen this approach mentioned before on similar questions and am wondering if it is valid?

  • Related: https://math.stackexchange.com/questions/4969779/uniform-blooming-times-of-two-flowers – Henry Oct 16 '24 at 12:34

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Yes, this approach is well known, though I have seen it more commonly used for people trying to meet at a particular time, each turning up uniformly over a time interval at random and then waiting some given period of time. One approach is to draw a square or rectangle with some diagonal lines and work out the area between the lines (cases where the two people coincide) as a proportion of the total area of the square/rectangle (when they might arrive). Chance of meeting in a bar illustrates this type of question.

Your version is slightly different, looking at discrete days rather than continuous times, but the same approach works with counting rather than areas. For example in the following picture, there are $100$ dots for when the two flowers might blossom, and the $43$ dots between the lines are cases where they might then both be in bloom together, so giving your $\frac{43}{100}$ probability.

100 dots and two diagonal lines

Henry
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  • A variation on the counting at the end is to note that the excluded top left / bottom right are both triangles. Hence we we can instead count these inadmissible cases as $\frac82(8+1)+\frac62(6+1)=36+21=57$, which indeed leaves $43$ admissible cases. (This isn't much better than counting by eye for this particular system, but allows for straightforward generalization when the "blooming seasons" for each flower are scaled up.) – Semiclassical Oct 14 '24 at 02:19