I'm not sure how an explanation is wrong

[full willow]

I got Cameron right of the attached, but after reading Richard's statement a few times I still don't get it. Could someone explain why Richard is wrong to me in some other way?

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[vocal summit]

solution just says "expected value $\neq$ probability thus is false"

[fickle fiber]

!𒐪 ɹɐupoɯ⇂ㄥ8𝟝 𒐪!

[full willow]

Can you explain to me the difference?

As in make your explanation longer.

[sonic geode]

This might be a little above your pay grade but the reason its confusing in this instance is because of how closely related probability of success and expected values are in a binomial distribution (as it is here). If probability for egg breaking is p and you have n eggs then the expected number of egg breaks happens to be np.

The expected number of broken eggs in a carton of 6 eggs is 0.12 that means the probability of getting a broken egg is 0.02. As it so happens the probability of getting exactly one broken egg in a carton of 6 is 6C1(0.02)^1(0.98)^5 roughly 0.10 or 10% ish...

[sonic geode]

Its better if you can more clearly commit to intuition that expected number and probability very different but related concepts. What do you understand of it right now explain it in your own words

[full willow]

Sorry connection dropped. 12% of 1 egg broken per pack of 6 eggs over a high # of packages, ie 200.

[hollow umbra]

Well, the given thing is E(number of broken eggs per carton), not P(carton contains a broken egg).

In fact, we can calculate it.

Suppose X ~ Bin(n = 6, p) is the number of broken eggs per carton. Then:
E(X) = np
So, p = E(X)/n. Thus, the probability that a carton contains a broken egg is:
P(X > 0) = 1 - (1 - p)^n = 1 - (1 - E(X)/n)^n = 1 - (1 - 0.12/6)^6 ≈ 0.11

[full willow]

So since the probability is < than the expected # of eggs then there's like 1% chance of getting a 2nd egg broken?

[hollow umbra]

No, that doesn't mean anything.

Those things aren't connected.

Expected value and prooability are fundamentally different. They don't even have the same unit in general.

[full willow]

So to make it as simple as possible , expected value is like 3 eggs broken 2% of the time and probability is you'll have about 1 egg broken per 8 packages?

[hollow umbra]

Not sure what you mean.
If X is the number of broken eggs in a pack, then E(X) is the average number of broken eggs in a pack and P(X > 0) is the probability of at least one broken egg in a package.
In general E(X) and P(X > 0) are not connected.

[full willow]

I was trying to generalize the difference between the 2.

+thank @hollow umbra

[flint gull]

@full willow has given 1 rep to @hollow umbra

[hollow umbra]

You're welcome!

I recommend recalling some formulas about E(X) and Var(X) in general.

And you can do some exercises on binomial distribution if you want.

[full willow]

Thanks but the mathematics prerequisite for https://www-teluq-ca.translate.goog/site/etudes/offre/prog/programme-court-en-science-des-donnees/?_x_tr_sl=fr&_x_tr_tl=en&_x_tr_hl=en-US doesn't touch statistics so I'm doing the minimum for precalculus right now.

[hollow umbra]

This is probability theory, not statistics.

You will get to statistics later, maybe next semester.

[full willow]

+close