Question about solution

To this question: https://brilliant.org/practice/probability-is-everywhere/?p=3

I get that since we don't gain any information, the probability space stays the same. What I don't understand is that since we know how many chocolates were taken out why can't we use that information to calculate the probability?

Like this: Friend takes a chocolate. Chance that it was cherry over the chocolate is 1/2. We then know that the chance of it being a cherry the second time is 4/9. Multiply the two, and you get 4/18 -> 2/9.

I guess there's some flaw in my logic, but I can't for the life of me figure it out.

#Combinatorics

Easy Math Editor

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Comments

You can: there is a $1/2$ chance that he took a cherry, in which case you have a $4/9$ chance to eat a cherry. And there's a $1/2$ chance that he took a chocolate, in which case you have a $5/9$ chance to eat a cherry. So the answer is still $1/2 \cdot 4/9 + 1/2 \cdot 5/9 = 1/2.$

Patrick Corn - 3 years, 8 months ago

what is the purpose of adding both cases together?

Andrew Turner - 3 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$