It's simple to solve if we can assume that the number of sweets of each type is independent from each other.
This problem is easy to define, but I'm not finding it easy to solve...
Take some bags of mixed sweets with three different flavours: cherry, lemon and apple.
The number of each type of sweet in each bag is normally distributed.
The means are:
cherry 10
lemon 15
apple 7
They all have a standard deviation of 3.
How can I calculate the probability that most of the sweets in any given bag will be cherry?
Any help would be very greatly appreciated, however general or specific.
I suggest let X=distribution of number of cherry sweets, Y for lemon and Z for apple. We want to find P(X-Y>0) and P(X-Z>0)
X-Y has mean 10-15=-5 variance 9+9=18 so standard deviation root 18. X-Z has mean 10-7=3 and same standard deviation.
Thanks for your response- yes they are independent from each other.It's simple to solve if we can assume that the number of sweets of each type is independent from each other.
That is very helpful- thanks so much.
I now have
P(X-Y>0) mean -5 dev 4.242640687 prob 0.119296415
P(X-Z>0) mean 3 dev 4.242640687 prob 0.760249939
Please could you tell me how to combine these into a single probability for X>A AND X>Z (ie: how often cherries make up the majority of the bag)
I assume it is an operation I have to perform on the two continuous distributions, not their probabilities, but my attempts have failed so far.
Ok- thanks so much for looking at this. Im sure its close.
I have just run a larger number of sims and can confirm that multiplying is out by around 0.02, while the two distributions are out by only 0.002 at most.