probability

[final bridge]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[cinder walrus]

Conditional probabilities

In order to get the probability of two events that are unrelated to each other, multiply both probabilities with each other

[zenith stone]

2/5 x 1/3?

[cinder walrus]

We take the probability that rebecca does not pass
$\frac{2}{5}$
and we multiply it with the probability that henry does pass
$\frac{1}{3}$

So we get:
$\frac{2}{5} * \frac{1}{3} = \frac{2}{15}$

[weak grotto]

muerte

[zenith stone]

Thank you

[cinder walrus]

no problem 😄

[cinder walrus]

This is a bit harder, because we have to consider multiple cases:

1. of all, when you have multiple events that furfill the condition, you just add the probability up
2. for each event, the same conditional probabilities from earlier matter
   In your case, there are just two situation where he wins exatly one game:

(Win, Lose) and (Lose, Win)

So lets get the probabilities for each:

(Win x Lose):
$\frac{8}{11} * \frac{3}{11} = \frac{24}{121}$

Now for (Lose x Win): (this has the same probability as Lose x Win)

$\frac{3}{11} * \frac{8}{11} = \frac{24}{121}$

Now we add it up

$2\frac{24}{121} = \frac{48}{121} $

[weak grotto]

muerte

[zenith stone]

ohh that make sense thanks

[cinder walrus]

no problem 🙂

[zenith stone]

+close