Compound probability: you choose a tile at random from a bag containing 2 tiles with x, 6 tiles with y, and 4 tiles with z. you pick a second tile without replacing the first. find each probability.

18. p(x then y) 19. (both y) 20. p(y then x) 21. (p(z then x)

22. p(both z) 23. p(y then z)

you don't have to answer all of them, i just want someone to solve like two and explain how they did it so i can do the rest on my own.

First, we must count the amount of tiles.

Amount = 2 + 6 + 4

A = 12 tiles

Then let's to the Example 18:

P(X then Y) = P(x) × P(y)

We have 2 lites x

P( x then y) = 2 / 12 × P(y)

We have 6 lites y

And as one lites X was choosed:

P(y) = 4 / 11

P(x then y) = (2/12) × ( 4 /11)

= 6.06%


Example 19:

P(both y) = P(y)' × P(y)

We have 4 lites y

P(y)' = 4/12

Now us have 3 lites y

P(y) = 3 / 11

P(both y) = (4/12)×(3/11)

= 9.09%


Example 20:

P(y then x) = P(y) × P(x)

We have 4 lites y

P(y) = 4 /12

Now us have 11 lites on the bag

P(x) = 2 / 11

P(y then x) = (4/12)×(2/11)

= 6.06%


Example 21:

P(z then x) = P(z) × P(x)

We have 6 lites z

P(z) = 6 / 12

Now us have 11 on the bag

P(x) = 2/11

P(z then x ) = (6/12)×(2/11)

= 9.09%


Example 22:

P(both z) = P(z)'×P(z)

P(z)' = 6/12

Now us have 5 lites Z

P(z) = 5/11

P(both z) = (6/12)×(5/11)

= 22.72%


Last example:

P(y then z) = P(y)×P(z)

We have 4 lites y

P(y) = 4/12

Now us have 11 lites on the bag

P(z) = 6/11

P(y then z) = (4/12)×(6/11)

= 18,18%

This would be your answers.

Hope this helps

I’m sorry i can’t u