KingGeorgeIII
contestada

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.

Respuesta :

David1993
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

Otras preguntas