Probability Question

[wraith jolt]

You roll a slot machine with 10% chance of success 11 times, then roll a slot machine with 5% chance of success 6 times. What's the probability of getting exactly 7 successes

[wraith jolt]

The idea I have for it is just to convert it into one machine
11 times with 0.1 expected value + 6 times with 0.05 expected value => 1.4 expected value with 17 times

So a machine with (140/17)% success rate per roll should have the same expected value as the 2 machines listed in the question combined but I dont think would have the same probability of getting exactly 7 successes (for example, a machine that does the first 3 rolls at 100% success rate and the last 3 rolls at 0% success rate versus a machine that does 6 rolls at 50% success rate have the same expected value but different probability to get 3 successes or 4 successes)

[fresh field]

I mean if nothing else you can calculate with binomial distribution for all possible scenarios like 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1, 7 and 0

Thats sounds managable

[wraith jolt]

Damn I only learnt the basics for that not how to calculate it for more complex questions

[fresh field]

it came out to about 1.9446+10^-4 for me

either way just used this and summed it up in my calculator

[wraith jolt]

Hmm alright

I was thinking of solving it in a middle school level kind of way

[fresh field]

Thats a high school level prob

I mean if you encounter something like this than you probably learned this?

[fresh field]

Have you not seen this before?

[fresh field]

This seems like an oversimplification and will probably yield wildly different numbers

okay maybe not that wild

it was about 10% off

Either way you cant combine them

[fresh field]

And even if you did I still had to use that

[wraith jolt]

Nah I just think of the question myself

I want to make my own math exam and is trying to think of a problem for probability

I've just entered high school this year lol

[fresh field]

See, well the binomial distribution is fairly simple

[wraith jolt]

If it's just 1 thing then I can just subtract the probability of getting atleast 7 by the probability of getting atleast 8

[fresh field]

its just the probability of success^successes and the probability af failure^failures times the ways it can happen

[fresh field]

I mean P(atleast 7) - P(atleast 8) = P(7)

but how do you get those probabilites?

[spice sonnet]

Good job!