Five Different Prize Machines

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Sekuiya

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Feb 24, 2015
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First of all, I stumbled on this problem and don't even know how to start calculating this, so, if only for that, it would be a good start.

Basically, we have 5 different prize machines, and every single one has the prize you desire but with different %.
For example, machine A has 4% to give out the desired prize, machine B has 1%, machine C has 2%, machine D has 3% and machine E has 3%.
If you have a go at the 5 different machines, what's the probability of getting the desired prize?

Hope you can help me with my problem.
Thanks in advance.
 
Sekuiya said:
First of all, I stumbled on this problem and don't even know how to start calculating this, so, if only for that, it would be a good start.

Basically, we have 5 different prize machines, and every single one has the prize you desire but with different %.
For example, machine A has 4% to give out the desired prize, machine B has 1%, machine C has 2%, machine D has 3% and machine E has 3%.
If you have a go at the 5 different machines, what's the probability of getting the desired prize?

Hope you can help me with my problem.
Thanks in advance.
Click to expand...
The probability of succeeding is one minus the probability of failing. Assuming the probabilities are all independent, what is the probability of failing each time.
 
I'm sorry Ishuda, I didn't quite understand where you wanted to go with your suggestion, probably because I tried studying a bit and reach this conclusion:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

In my case, having in mind that they are all independent, does that mean that the result of P(desired item) is the ΣP(desired itemi) ?
But in that case, what would happen if there were 100 machines with 1% in every one? Does that mean the probability of getting the desired item is 100%? I think I'm missing something here.
 
Sekuiya said:
I'm sorry Ishuda, I didn't quite understand where you wanted to go with your suggestion, probably because I tried studying a bit and reach this conclusion:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

In my case, having in mind that they are all independent, does that mean that the result of P(desired item) is the ΣP(desired itemi) ?
But in that case, what would happen if there were 100 machines with 1% in every one? Does that mean the probability of getting the desired item is 100%? I think I'm missing something here.
Click to expand...
Let's do an example:
Three machines
First machine 25% to win prize implies 75% to not win
Second Machine 35% to win and 65% not to win
Third Machine 20% to win 80% to lose

Assuming each machine is independent of the others, the chance of losing on all three machines is 0.75 * 0.65 * 0.80 = 0.39 or 39%. The opposite of losing all three times is winning at least one time [maybe 2 or 3 times but at least one]. Thus the chance of winning at least once 1-.39 = .61 or 61%.
 
Oh, oh, I think I get it now, thanks!

So, in my case it would be 0.96 * 0.99 * 0.98 * 0.97 * 0.97 = 0.8763 , 1 - 0.8763 = 0.1237, P(desired item) = 0.1237 = 12.37%.

Your first suggestion now makes sense, sorry for not understanding it the first time.
Thanks for all the help so far!
 
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