Answer:
1) a) Check Below b) 9 2) Dependent b) Dependent 3)[tex]P(Pair \:of \:Pants \:or\:Blue\:Clothing)=\frac{10+4+8}{26}=\frac{22}{26}=\frac{11}{13} =0.85[/tex]
Step-by-step 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:
[tex]P(A\cap B) = P(A)\times P(B)[/tex]
Rewriting the Chart
[tex]Home\\\\P(Win)=0.2\\P(Loss)=0.6\\P(Total)=0.8[/tex]
[tex]Away:\\P(Win)=0.05\\P(Loss)=0.1\\P(Total)=0.15[/tex]
[tex]Total\\\\P(Win)= 0.25\\P(Loss)=0.7\\P(Total)=1.00[/tex]
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:
[tex]P(Blue) =\frac{18}{26}=\frac{9}{13}[/tex]
[tex]P(Pair \:of \:Pants \:or\:Blue\:Clothing)=\frac{10+4+8}{26}=\frac{22}{26}=\frac{11}{13} =0.85[/tex]
