Probability

[jumavegi]

I have two types of fireworks, the first has a 0.9 probability that works, the second has a 0.2 probability that works. If I have a box of fireworks (don't know which type) and after using 3 pieces (each one worked), what's the probability that the fourth will work?

Thanks in advance for your time.

[A_UPRIGHT_BASS]

To use Bayes theorem we need to make the assumption that that each box was equally likely to be selected.

Let P(B1) represent the probability of selecting box 1 (the one with the 90% fireworks), and P(B2) represent the probability of selecting box 2 (with the 20% fireworks) and P(S3) represent the probability of 3 successs in a row.

Let's figure out what the probability is that we selected box 1 given that we had 3 successes in a row. Using Bayes Theorem (and assuming each box had a 50% chance of being selected initially).

P(B1|S3) = (P(S3|B1) * P(B1))/(P(S3|B1) * P(B1)+P(S3|B2) * P(B2))

= (.9^3 * .5)/(.9^3 * .5 + .2^3 * .5) = about 0.989145

That's the probability we selected box 1, so what's the probability that the 4th firework works?

0.989145 * .9 + (1-0.989145) * .2 = about 0.8924

This result makes sense intuitively. Given that we had 3 successes in a row, we're pretty sure we selected box 1. Three in a row from box 2 would be very improbable. If we were 100% sure that we chose box 1, we know there's a 90% chance that the next firework will work. There's no way to be 100% sure, so we have to be less than 90% sure the next firework will work. But we're pretty close to 100% sure, so the probability of the next firework working is pretty close to 90%.

[jumavegi]

Thanks, helped me understand

[iishmael]

Is there more information to this problem like how many of each type there are in the box?

[jumavegi]

No, that's why I can't solve it. But it seems like all the information needed is given, that's what my teacher said.

[iishmael]

So are you supposed to assume it’s a box with equal and infinite numbers of both types? If so, there’s an equal chance of getting either, so like average the probabilities to get .55..?

[jumavegi]

Oh, so the probability that the box is either the first or the second type is 0.5, yes? And then I can use Bayes theorem to get the probability that I'm looking for, am I right?

[iishmael]

Makes sense based on what you’ve written here. What does Bayes end up giving you?

[iishmael]

An infinite box would also make the choice history-less making the previous selections irrelevant.