Question

REAL ANSWERS PLEASE...

1. For dinner, Javier will choose one of three fish options and one of three side dishes. His choices for fish are salmon, trout, and halibut. His choices for side dishes are fries, cooked carrots, or coleslaw.
(a)Draw a tree diagram for the sample space of all dinner combinations.
(b)How many choices for dinner combinations does Javier have?
Answer:

2. The table shows the probabilities of winning or losing when the team is playing away or is playing at home.
Home: Away: Total:
Win: .2 .05 .25
Loss: .6 .1 .7
Total: .8 .15 1.00

(a) Are the events “winning” and “playing at home” independent? Explain why or why not.
(b) Are the events “losing” and “playing away” independent? Explain why or why not.
Answer:

3. A box contains 4 black shirts, 8 blue shirts, 4 black pants, and 10 blue pants. Determine the probability of randomly selecting a blue piece of clothing or a pair of pants. Use to explain your answer.
Answer:

Answer 1

1) a) Check Below b) 9 2) Dependent b) Dependent 3)
`P(Pair \:of \:Pants \:or\:Blue\:Clothing)=(10+4+8)/(26)=(22)/(26)=(11)/(13) =0.85`

Explanation:

1)

Let's match the possibilities:

Salmon, Fries Salmon, Cooked Carrots Salmon, Coleslaw

Trout, Fries Trout, Cooked Carrots Trout, Coleslaw

Halibut, Fries Halibut, Cooked Carrots Halibut, Coleslaw

b) 9 choices

__3_ * __3___=9

2) If the Probability of "Winning" (A) happens independently from the Probability of "Playing at Home" (B) and the result of P(A) does not influence P(B). We can calculate the probability of Winning and Playing at Home:

`P(A\cap B) = P(A)* P(B)`

Rewriting the Chart

`Home\\\\P(Win)=0.2\\P(Loss)=0.6\\P(Total)=0.8`

`Away:\\P(Win)=0.05\\P(Loss)=0.1\\P(Total)=0.15`

`Total\\\\P(Win)= 0.25\\P(Loss)=0.7\\P(Total)=1.00`

a) Since the Total Events were not found by the product of "winning" and "playing at home" we cannot say these are independent events.

Dependent Event

b) Dependent Events

Let's remember the number of matches goes decreasing since the first one, and that's what makes them dependent in probability, in other words, the denominator (sample space) decreases in the product.

3a) Blue Piece of Clothing. The Clothing is made up by shirt and pant

We have 12 shirts and 14 pair of pants, totalizing 26 clothing.

But the Blue ones are 8 shirts and 10 pants,

Then, let's add the the amount of 4 black pants to the numerator and then add it, since we are including besides the blue clothing, the black pants over the the whole number of clothing:

`P(Blue) =(18)/(26)=(9)/(13)`

`P(Pair \:of \:Pants \:or\:Blue\:Clothing)=(10+4+8)/(26)=(22)/(26)=(11)/(13) =0.85`

Answer 2

1.) - Salmon Trout Halibut
/ | \ / | \ / | \
Fries Carrots Slaw F C S F C S
-There is 9 different options, assuming that since its a combination problem and not a sequence problem, order doesn't matter.