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.

Note by Andrew Turner
8 months, 2 weeks ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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 $$1/2 \cdot 4/9 + 1/2 \cdot 5/9 = 1/2.$$

- 8 months, 2 weeks ago

what is the purpose of adding both cases together?

- 8 months, 2 weeks ago