Weekend HW = AP Classroom = Quiz and HW Grades!

Weekend HW = AP Classroom

1.) HW = AP Classroom Assignment = Unit 4 Progress Check MC A

• Your score = quiz grade, but also....
• Show your work (where applicable) and record your answers for each question on separate paper--this will be checked Monday as a homework grade as well!
• (This is also a backup in case your assignment does not submit for some reason, I can grade the paper stuff)
• Then, grade yourself--show corrections for any questions you got wrong!
• Here are some tips for a few of the tricky ones....
• #3: This question deals with simulation, something we have not discussed--no biggie though.
• For a simulation you want the random numbers you use to match the given probability.
• For example, if I wanted to use a simulation with random numbers for coin flips, I know the probability of getting heads (or tails) is 0.5, so I would want 5 out of the 10 (single digit) integers to represent heads, and 5 to represent tails
• So I could simulate with 0,1,2,3,4 = heads, 5,6,7,8,9 = tails
• Another example: let's say the probability of winning a game is 0.7. To simulate playing this game using digits 0-9 I would want 70% or 7 of the digits to represent winning, so:
• Winning = 0,1,2,3,4,5,6
• Losing = 7,8,9,10
• #10: This question asks if we have mutually exclusive events--for events to be mutually exclusive (aka disjoint), this means "both can't happen," or mathematically speaking, the "probability of A and B must = 0."
• So, for this question we need the general addition rule:
• We know:
• P(Brown Hair or Brown Eyes) = P(Brown Hair) + P(Brown Eyes) - P(both)
• We are given the probability of brown hair, brown eyes, and brown hair or brown eyes...
• Substitute these values into the equation above, solve for the P(both brown hair and brown eyes), and see what you get!
• If the P(both) = 0, that means "both" can't happen and they're disjoint; if you get a number, that means "both" can happen and they are not disjoint
• #11: Nothing tough here, but the answer is written as a fraction--just multiply your numerators and your denominators to match the MC option
•  #12: For this one I would draw a Venn diagram; you are given 3 of the 4 numbers for the Venn diagram...
• Again, we're back to disjoint/mutually exclusive...
• We need to find the probability of "both," or for the middle in our Venn diagram...
• If this probability (in the middle) = 0, events are disjoint/mutually exclusive
• If this probability (in the middle) is not 0, events are not disjoint/mutually exclusive
• "joint probability" = "probability of both" = "probability in the middle of the Venn diagram"
• #15: This will likely be the toughest one; this question is very formulaic, and is based on the conditional probability formula we learned today.....
• First, write the conditional notation (in words) for the given 37.5%--this represents the probability of "___ given ____."
• Now, consider the conditional probability formula....
• P(A given B) = P(A and B)/(P/B)
• In this example we have the numbers for the left side, "A given B," and for the denominator...
• Sub in those values and then solve for the numerator, "A and B"

2.) Bonus Assignment = more practice identifying conditional probability!

• Read each question below
• Determine which questions are conditional probability; for the ones that are, please write the correct conditional notation "in words"
• (Same as last night's directions)
• You can write all answers on separate paper or in your notes....
• This is due Tuesday (I'l have paper copies in class Monday).
• Your extra credit will be based on accuracy/correctness, not just effort. 

Monday's HW:

• Page 363: 11, 21, 23a, 35/37, 39/41
• 35 and 37 use the same context, that's why they're listed that way (35/37). Same for 39 and 41.
• We will learn how to do 35/37 and 39/41 in class on Monday, but you could do 11, 21, and 23a based on the conditional probability formula we developed in class.