Advanced Math Topics Chapters 4 and 5 Review. 1) A family plans to have 3 children. What is the probability that there will be at least 2 girls? (assume.

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Advanced Math Topics Chapters 4 and 5 Review

1) A family plans to have 3 children. What is the probability that there will be at least 2 girls? (assume that there is equal probability of a baby boy or girl) First, find the sample space. GGG GGB GBG GBB BBB BBG BGB BGG How many have at least 2 girls? 1 st Child2 nd Child3 rd Child P(at least 2 girls) = 4/8 = 1/2

Sample space of multiple events = M N The first event has a sample space of M The second event has a sample space of N This formula extends to more than two events done in a sequence.

In a European nation, license plates have 3 letters followed by 2 digits. If the first digit cannot be 0, how many different license plates can be made if… a) Repetitions are allowed. 26 x 9 x10 =1,581,840 possibilities b) Repetitions of letters are not allowed. 26 x25 x24 x9 x10 =1,404,000 possibilities

Al, Bob, Carey, Dee, and Eric have 5 front row seats at a concert. They each can sit in any of the 5 seats. a) How many seating possibilities are there? A person cannot sit in two seats, thus it is implied that there are no repetitions. 5 x120 possibilities 4 x3 x2 x1 =

Permutation:An arrangement of distinct objects in a Particular order Permutations are Particular Picks. The number of permutations of “n” things picked “r” at a time is… n P r = n! (n – r)! (order matters)

Al, Bob, Carey, Dee, and Eric are the only entries in a drawing for two front-row tickets to a concert. How many seating possibilities are there if two of them are picked? How does this change the solving process? 5 x20 possibilities 4= There are 5 people being picked in a particular order 2 at a time. 5 pick 2 = 5 P 2 = 5! (5 – 2)! = 5! 3! = 5x4x3x2x1 3x2x1 = 20 There are two ways to solve this and it may help explain the permutation formula.

You are putting 2 identical clown dolls and 4 identical leprechaun dolls on the shelf. In how many different ways can this be done? Does this work? 6 x5 x4 x3 x2 x1 =6! =720 ways This means that in the first spot you have 6 choices, then you have 5 choices, etc. But some of the dolls are identical so this number is inflated! Switch the 1 st and 3 rd leprechaun to get this… They are the same arrangement but the solution above counts both of them. Thus, we need a new way to solve this to get fewer possibilities than 720. The number of permutations of “n” things of which “p” are alike, “q” are alike, or “r” are alike, and so on, is… n! p!q!r! where p + q + r…. = n. Using this formula: 6! 4!2! = 6x5x4x3x2x1 (4x3x2x1)(2x1) = 15 This is the same solving strategy for… How many ways can you rearrange the letters of the word “mathematics”?

7) You are selecting a 4-person committee from the 50 Governors. There are 38 male Governors and 12 female Governors. How many different committees can be selected if you want to select 3 women and 1 man? Answer: 8,360 Process: 12 C 3 x 38 C 1 = Teams or Clubs, use combinations.

Mathematical Expectation: The amount of money expected to be won or lost in an event or series of events. Mathematical Expectation Formula = m 1 p 1 + m 2 p 2 + m 3 p 3 + … + m n p n The amount earned (or lost) if an event occurs times the probability of that event. Note: If money will be lost when the event occurs, then m is negative.

9) Two dice are rolled. If you roll a sum greater than 9, you win $4. Otherwise, you lose $1. What is your mathematical expectation? Round to the nearest penny. Answer: -$0.17 Process: 4,6 6, 4 5,5 5,6 6,5 and 6,6. The probability of rolling a sum greater than 9 is 6/36. Thus, the probability of the other outcomes is 30/36. (6/36)($4) + (30/36)(-$1) =

Odds in favor of an event = p : q The number of favorable outcomes. The number of unfavorable outcomes.

3) If the probability of the New York Mets winning a particular game is 3/7, what are the odds in favor of the Mets winning the game? Answer: 3 : 4 Process: 3 favorable outcomes to 4 unfavorable outcomes.

Addition Rule: If A and B are any events… p(A or B) = p(A) + p(B) – p(A and B) Example: p(red card or a Jack) =p(red) + NOTES p(Jack) -p(red jack) = 26/52 +4/52 -2/52 =28/52 = 14/26 = 7/13

Answer: 96/631 Process: ( )/total # of students 192/ ) You select a student at random. Find the probability that you selected a student whose favorite soda is Root Beer. Leave as a reduced fraction. Coke Pepsi Mt. Dew Root Beer Juniors Seniors Favorite Sodas at SRVHS

Answer: 111/313 Process: P(Coke | senior) = P(coke and senior)/P(senior) = 222/626 4) You select a student at random. Find the probability that you selected a student whose favorite soda is Coke given that the student is a senior. Leave as a reduced fraction. Coke Pepsi Mt. Dew Root Beer Juniors Seniors

Multiplying Independent Events Independent Events: Events whose probabilities are the same no matter what happens with the other event If A and B are Independent Events, then… p(A and B) = p(A) p(B) Example: You flip two coins and roll two dice. Find the probability of flipping a heads, a tails, rolling an even and anything other than a 4. ½ ½ ½ 5/6 = 5/48 The events are independent!

Multiplying Conditional Events Conditional Events: Events whose probabilities depend upon what happens in previous events If A and B are Conditional Events, then… p(A and B) = p(A) p(B | A) Example: You select 2 cards from a deck of 52. You select the cards one at a time. Find p(diamond and a black ace) 13/52 2/51 = = p(diamond) p(black ace | diamond) 1/4 2/51 =2/204 = 1/102

HW P. 231 #1-5, 11, 16 P. 282 #2, 8, Check answers