10000000000Hello
I'd like to understand the process of how to calculate the probability of the following problem....
I have a digital picture frame that shows 2 pictures per day (morning and afternoon). It has a selection of 10 pictures and doesn't remember any previous choices i.e. it could show the same picture many times. Over the course of 5 days, what is the probability of seeing all 10 photos.
My working so far has been to start with 2 chances and 2 photos. So that means 4 possible combinations of photos with 2 combinations that mean I see both i.e.
1, 2 <- Winner (I see both photos)
2, 2
2, 1 <- Winner (I see both photos)
1, 1
So that suggests I have a 50% chance of seeing both photos in 1 day (4 possible outcomes, 2 winning outcomes = 2/4 = 0.5).
If I scale that up to 3 photos with 3 chances (i.e. morning, afternoon, morning of the next day) I have 27 (3^3) possible combinations and 6 (3 factorial) winners - which gives 6/27 = 22% chance
So my thinking here is that if I scale the number of photos with the number of chances I should arrive at the correct probability. So for 5 days with 2 photos per day that's 10 chances, and 10 photos.
So on that basis it's 10^10 = 10000000000 possible combinations, and 3628800 winning outcomes (10 factorial), which gives 0.036288% (3628800 / 10000000000)
Is that the right way to approach this or am I completely on the wrong track? If I am on the right track I'm not quite sure why!
The other way I was looking at it was to say for any one change of picture there is an equal chance of picking any of the 10 photos i.e.
1 / 10 = 0.1 = 10%
So if I take 10 chances and multiply the probabilities I get
0.1*0.1*0.1*0.1*0.1*0.1*0.1*0.1*0.1*0.1 = 0.0000000001
But that feels like it's telling me what is the chance of picking the same photo each time, and I don't know how to translate that into the chance of picking all photos.
Any ideas?
Cheers
David
I'd like to understand the process of how to calculate the probability of the following problem....
I have a digital picture frame that shows 2 pictures per day (morning and afternoon). It has a selection of 10 pictures and doesn't remember any previous choices i.e. it could show the same picture many times. Over the course of 5 days, what is the probability of seeing all 10 photos.
My working so far has been to start with 2 chances and 2 photos. So that means 4 possible combinations of photos with 2 combinations that mean I see both i.e.
1, 2 <- Winner (I see both photos)
2, 2
2, 1 <- Winner (I see both photos)
1, 1
So that suggests I have a 50% chance of seeing both photos in 1 day (4 possible outcomes, 2 winning outcomes = 2/4 = 0.5).
If I scale that up to 3 photos with 3 chances (i.e. morning, afternoon, morning of the next day) I have 27 (3^3) possible combinations and 6 (3 factorial) winners - which gives 6/27 = 22% chance
So my thinking here is that if I scale the number of photos with the number of chances I should arrive at the correct probability. So for 5 days with 2 photos per day that's 10 chances, and 10 photos.
So on that basis it's 10^10 = 10000000000 possible combinations, and 3628800 winning outcomes (10 factorial), which gives 0.036288% (3628800 / 10000000000)
Is that the right way to approach this or am I completely on the wrong track? If I am on the right track I'm not quite sure why!
The other way I was looking at it was to say for any one change of picture there is an equal chance of picking any of the 10 photos i.e.
1 / 10 = 0.1 = 10%
So if I take 10 chances and multiply the probabilities I get
0.1*0.1*0.1*0.1*0.1*0.1*0.1*0.1*0.1*0.1 = 0.0000000001
But that feels like it's telling me what is the chance of picking the same photo each time, and I don't know how to translate that into the chance of picking all photos.
Any ideas?
Cheers
David