Probability Probability Day 4

Independent Practice Topic 3 packet pg

Opening Routine

Opening Routine # 2

Opening Routine # 3

Answers: 1- 4/25 = 0.16 or 16 % 2- 17/50 = 0.34 or 34 % 3- Choice A ( the more trials the closer to theoretical probability)

Conditional Probability Conditional probability is the probability of an event occurring, given that another event has already occurred. Conditional probability restricts the sample space. The conditional probability of event B occurring, given that event A has occurred, is denoted by P(B|A) and is read as “probability of B, given A.” We use conditional probability when two events occurring in sequence are not independent. In other words, the fact that the first event (event A) has occurred affects the probability that the second event (event B) will occur.

Conditional Probability Formula for Conditional Probability Better off to use your brain and work out conditional probabilities from looking at the sample space, otherwise use the formula.

Conditional probability problems can be solved by considering the individual possibilities or by using a table, a Venn diagram, a tree diagram or a formula. e.g. There are 2 red and 3 blue counters in a bag and, without looking, we take out one counter and do not replace it. The probability of a 2 nd counter taken from the bag being red depends on whether the 1 st was red or blue. Conditional Probability

e.g. 1. The following table gives data on the type of car, grouped by gas consumption, owned by 100 people. One person is selected at random Female Male HighMediumLow 100 Total L is the event “the person owns a low rated car”

Female Low Male e.g. 1. The following table gives data on the type of car, grouped by gas consumption, owned by 100 people HighMedium One person is selected at random. L is the event “the person owns a low rated car” F is the event “a female is chosen”. 100 Total

Female Low Male e.g. 1. The following table gives data on the type of car, grouped by gas consumption, owned by 100 people HighMedium One person is selected at random. L is the event “the person owns a low rated car” F is the event “a female is chosen”. 100 Total

Female Low Male e.g. 1. The following table gives data on the type of car, grouped by petrol consumption, owned by 100 people HighMedium One person is selected at random. L is the event “the person owns a low rated car” F is the event “a female is chosen”. There is no need for a Venn diagram or a formula to solve this type of problem. We just need to be careful which row or column we look at. Find (i) P (L) (ii) P (F  L)(iii) P (F L) 100 Total

(i) P (L) = Solution: Find (i) P (L) (ii) P (F  L)(iii) P (F L) Female 733Male TotalHighMediumLow Low Probability that “the person owns a low rated car”

(i) P (L) = Solution: Find (i) P (L) (ii) P (F  L)(iii) P (F L) Female 73312Male HighMediumLow (ii) P (F  L) = The probability of selecting a female with a low rated car Total

(i) P (L) = Solution: Find (i) P (L) (ii) P (F  L)(iii) P (F L) Female 73312Male HighMediumLow (ii) P (F  L) = Total (iii) P (F L)  The probability of selecting a female given the car is low rated. We must be careful with the denominators in (ii) and (iii). Here we are given the car is low rated. We want the total of that column The sample space is restricted from 100 to 35.

(i) P (L) = Solution: Find (i) P (L) (ii) P (F  L)(iii) P (F L) Female 73312Male HighMediumLow (ii) P (F  L) = Total (iii) P (F L)  Notice that P (L)  P (F L) So, P (F  L) = P(F|L)  P (L) = P (F  L) Try this: P(F), P( L/F) and P( F  L)= P(L/F)* P(F)

(i) P (L) = Solution: Find (i) P (L) (ii) P (F  L)(iii) P (F L) Female 73312Male HighMediumLow (ii) P (F  L) = Total (iii) P (L F)  Notice that P (L)  P (F L)= P(F) x P( L/F) So being a female and owning a low rated vehicle are independent events!!! = P (F  L)

Ex. 2: I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution:LetR be the event “ Red flower ” and F be the event “ First packet ” R F Red in the 1 st packet

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F 8 Red in the 1 st packet LetR be the event “ Red flower ” and F be the event “ First packet ”

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F Blue in the 1 st packet 8 LetR be the event “ Red flower ” and F be the event “ First packet ”

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F Blue in the 1 st packet 812 LetR be the event “ Red flower ” and F be the event “ First packet ”

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F Red in the 2 nd packet 812 LetR be the event “ Red flower ” and F be the event “ First packet ”

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F Red in the 2 nd packet LetR be the event “ Red flower ” and F be the event “ First packet ”

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F Blue in the 2 nd packet LetR be the event “ Red flower ” and F be the event “ First packet ”

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F Blue in the 2 nd packet LetR be the event “ Red flower ” and F be the event “ First packet ”

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F Total: LetR be the event “ Red flower ” and F be the event “ First packet ”

e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. 45 Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F Total: LetR be the event “ Red flower ” and F be the event “ First packet ”

45 e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: R F LetR be the event “ Red flower ” and F be the event “ First packet ”

45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: P (R  F) = LetR be the event “ Red flower ” and F be the event “ First packet ”

45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: P (R  F) = LetR be the event “ Red flower ” and F be the event “ First packet ”

45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: P (R  F) = LetR be the event “ Red flower ” and F be the event “ First packet ”

45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: P (R  F) = P (R F) = 8 LetR be the event “ Red flower ” and F be the event “ First packet ”

45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: P (R  F) = P (F) =P (R F) = 8 LetR be the event “ Red flower ” and F be the event “ First packet ”

45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: P (R  F) = P (F) =P (R F) = 8 LetR be the event “ Red flower ” and F be the event “ First packet ”

45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: P (R  F) = P (F) =P (R F) = 8 LetR be the event “ Red flower ” and F be the event “ First packet ”

45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: P (R  F) = P (F) =P (R F) = 8 P (R  F) = P(R|F)  P (F) So, P (R F)  P (F) = LetR be the event “ Red flower ” and F be the event “ First packet ”

The table shows the results of a class survey. Conditional Probability The condition female limits the sample space to 14 possible outcomes. Of the 14 females, 8 own a pet. Therefore, P(own a pet | female) equals yesno female86 male57 Do you own a pet? 14 females 12 males Find P(own a pet | female)

The table shows the results of a class survey. Conditional Probability The condition male limits the sample space to 15 possible outcomes. Of the 15 males, 7 did the dishes. Therefore, P(wash the dishes | male) is females 15 males Find P(wash the dishes | male) Did you wash the dishes last night? yesno female76 male78

Or Probabilities with Events That Are Not Mutually Exclusive If A and B are not mutually exclusive events, then P(A or B) = P(A) + P(B) – P(A and B). Using set notation,

Example 1 Toss a fair coin twice. Define A: head on second toss B: head on first toss HT TH TT 1/4 P(A|B) = ½ P(A|not B) = ½ P(A|B) = ½ P(A|not B) = ½ HH P(A) does not change, whether B happens or not… A and B are independent ! 1/4

Example 3: Two Dice Toss a pair of fair dice. Define A: red die show 1 B: green die show 1 P(A|B) = P(A and B)/P(B) =1/36/1/6=1/6=P(A) P(A|B) = P(A and B)/P(B) =1/36/1/6=1/6=P(A) P(A) does not change, whether B happens or not… A and B are independent !

Day 3 Review

Example: Additive Rule Example: Suppose that there were 120 students in the classroom, and that they could be classified as follows: BrownNot Brown Male2040 Female30 A: brown hair P(A) = 50/120 B: female P(B) = 60/120 P(A  B) = P(A) + P(B) – P(A  B) = 50/ / /120 = 80/120 = 2/3 P(A  B) = P(A) + P(B) – P(A  B) = 50/ / /120 = 80/120 = 2/3 Check: P(A  B) = ( )/120 Check: P(A  B) = ( )/120

Example: Two Dice A: red die show 1 B: green die show 1 P(A  B) = P(A) + P(B) – P(A  B) = 6/36 + 6/36 – 1/36 = 11/36 P(A  B) = P(A) + P(B) – P(A  B) = 6/36 + 6/36 – 1/36 = 11/36

A Special Case mutually exclusive, When two events A and B are mutually exclusive, P(A  B) = 0 and P(A  B) = P(A) + P(B). BrownNot Brown Male2040 Female30 A: male with brown hair P(A) = 20/120 B: female with brown hair P(B) = 30/120 P(A  B) = P(A) + P(B) = 20/ /120 = 50/120 P(A  B) = P(A) + P(B) = 20/ /120 = 50/120 A and B are mutually exclusive, so that

Example: Two Dice A: dice add to 3 B: dice add to 6 A and B are mutually exclusive, so that P(A  B) = P(A) + P(B) = 2/36 + 5/36 = 7/36 P(A  B) = P(A) + P(B) = 2/36 + 5/36 = 7/36

Example BlondeNot Blonde Male2040 Female30 A: male P(A) = 60/120 B: female P(B) = ? P(B) = 1- P(A) = 1- 60/120 = 60/120 P(B) = 1- P(A) = 1- 60/120 = 60/120 A and B are complementary, so that Select a student at random from the classroom. Define:

Example 2 A bowl contains five M&Ms ®, two red and three blue. Randomly select two candies, and define A: second candy is red. B: first candy is blue. m m m m m P(A|B) =P(2 nd red|1 st blue)= 2/4 = 1/2 P(A|not B) = P(2 nd red|1 st red) = 1/4 P(A|B) =P(2 nd red|1 st blue)= 2/4 = 1/2 P(A|not B) = P(2 nd red|1 st red) = 1/4 P(A) does change, depending on whether B happens or not… A and B are dependent!