Joint Probability of Independent Events

[Subliminal1]

Hello,
I'd like someone to double check my maths please, it wasn't the greatest at school and that was a few years ago!
If I am incorrect could you please tell me the correct answer as well as show me how to apply it so I can figure out similar problems myself. Thanks.

Yes, this is about Pokémon, but there is a lot of maths involved. Here goes:

PROBLEM 1:
For a Pokemon to be "shiny" the odds are 1/273 encounters.
Each encounter has the chance for "Pokemon A" to appear of 60% OR"Pokemon B" to appear of 40%.

The odds of having Pokemon A appear shiny are 1/455
I multiplied 1/273 by 6/10 to result in 1/455.
The odds of having Pokemon B appear shiny are 2/1365
I multiplied 1/273 by 4/10 to result in 2/1365.

PROBLEM 2:
For a Pokemon to be shiny the odds are 1/273 encounters.
5 Pokemon appear in each encounter and always 5, never more, never less. Each has the same odds of being shiny.

The chance that two Pokemon will be shiny in a single encounter are 1/74529
I squared/ multiplied by the power of 2, 273 to result in 1/74529

Applying this maths, the chance of all 5 to be shiny will be 273 to the power of 5?

PROBLEM 3:
I am unsure how to work this one out.
For a Pokemon to be shiny the odds are 1/273 encounters.
Each encounter has the chance for Pokemon A to appear of 60% OR Pokemon B to appear of 40%.
Pokemon A and B both have a 50% chance to be MALE and 50% chance to be FEMALE.

What are the chances of encountering a FEMALE Shiny Pokemon A?

Please let me know if my maths is correct, if it isn't please correct me and show me how to apply it too.

Many thanks!

 

[Sudharaka]

Hello,
I'd like someone to double check my maths please, it wasn't the greatest at school and that was a few years ago!
If I am incorrect could you please tell me the correct answer as well as show me how to apply it so I can figure out similar problems myself. Thanks.

Yes, this is about Pokémon, but there is a lot of maths involved. Here goes:

PROBLEM 1:
For a Pokemon to be "shiny" the odds are 1/273 encounters.
Each encounter has the chance for "Pokemon A" to appear of 60% OR"Pokemon B" to appear of 40%.

The odds of having Pokemon A appear shiny are 1/455
I multiplied 1/273 by 6/10 to result in 1/455.
The odds of having Pokemon B appear shiny are 2/1365
I multiplied 1/273 by 4/10 to result in 2/1365.

PROBLEM 2:
For a Pokemon to be shiny the odds are 1/273 encounters.
5 Pokemon appear in each encounter and always 5, never more, never less. Each has the same odds of being shiny.

The chance that two Pokemon will be shiny in a single encounter are 1/74529
I squared/ multiplied by the power of 2, 273 to result in 1/74529

Applying this maths, the chance of all 5 to be shiny will be 273 to the power of 5?

PROBLEM 3:
I am unsure how to work this one out.
For a Pokemon to be shiny the odds are 1/273 encounters.
Each encounter has the chance for Pokemon A to appear of 60% OR Pokemon B to appear of 40%.
Pokemon A and B both have a 50% chance to be MALE and 50% chance to be FEMALE.

What are the chances of encountering a FEMALE Shiny Pokemon A?

Please let me know if my maths is correct, if it isn't please correct me and show me how to apply it too.

Many thanks!

Hi Subliminal,

Considering each Pokemon is created independently from others your math is correct. For more information on why this works refer >>this<<. For the last question, the probability that the Pokemon is female is $\frac{1}{2}$, the probability that A appears is $\frac{60}{100}$ and the probability that it is shiny is, $\frac{1}{273}$. Considering these three events as independent the joint probability is, $\frac{1}{2}\times\frac{60}{100}\times\frac{1}{273}$.

 

[Subliminal1]

Hi Sudharaka,

I'm glad all those years at school actually got through to me - so "that" is what school is for!

But seriously, thank you for your reply so soon, it was well explained also.

Cheers
John/ Sub

 

[Sudharaka]

Hi Sudharaka,

I'm glad all those years at school actually got through to me - so "that" is what school is for!

But seriously, thank you for your reply so soon, it was well explained also.

Cheers
John/ Sub

You are very welcome. :)