Warm up:
Solve each system (any method)
2𝑥 − 𝑦 = 0
3𝑥 + 2𝑦 = 7
𝑥 = 4 − 2𝑦
2𝑥 + 4𝑦 = 8
𝐴𝑛𝑠𝑤𝑒𝑟: (1,2)
𝐴𝑛𝑠𝑤𝑒𝑟: 𝐼𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
W-up 11/4
• 1) Cars are being produced by two factories, factory 1
produces twice as many cars (better management) than
factory 2 in a given time. Factory 1 is know to produce
2% defectives and factory 2 produces 1% defectives. A
car is examined and found to be defective, what is the
probability it was produced by factory 1?
• Represent this mathematically
• Interpret the answer
• 2. evaluate b(7,4;.20) – show the steps
• 3. A fair coin is tossed 8 times, what is the probability of
obtaining at least 6 heads?
Answers: 1. 80% 2. 2.87% 3. 14.45%
8.3 EXPECTED VALUE
SWBAT compute expected values in addition to solving
application problems involving expected value.
Consider a coin flipping game: If heads shows, you
lose $1. If tails shows, you win $2.
• Let E be our expected value.
• 𝐸 = $2
1
2
+ −$1
1
2
= $. 5
• Where ½ is the probability of getting Heads or Tails
*So you expect to win an average of $.50 on each play.
Expected Value:
• S = Sample Space
• 𝐴1 , 𝐴2 , … 𝐴𝑛 = 𝑛 𝑒𝑣𝑒𝑛𝑡𝑠 𝑜𝑓 𝑆 𝑡ℎ𝑎𝑡 𝑓𝑜𝑟𝑚 𝑎 𝑝𝑎𝑟𝑡𝑖𝑡𝑖𝑜𝑛.
• 𝑝1 , 𝑝2 , … . 𝑝𝑛 = 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡𝑠 𝐴.
• 𝐴1 , 𝐴2 , … 𝐴𝑛 is assigned payoff 𝑚1 , 𝑚2 , … . 𝑚𝑛 .
• The Expected Value E corresponding to the payoffs is:
𝑬 = 𝒎𝟏 ∙ 𝒑𝟏 + 𝒎𝟐 ∙ 𝒑𝟐 … … . . +𝒎𝒏 ∙ 𝒑𝒏
Steps to compute E:
• Partition “S” into the “A” events.
• Determine the probability of each event (Sum of
probabilities should = 1).
• Assign payoff values “m”.
• Calculate.
Compute the expected value:
Outcome
𝒆𝟏
𝒆𝟐
𝒆𝟑
𝒆𝟒
Probability
1/3
1/6
1/4
1/4
• SS: {𝑒1 , 𝑒2 , 𝑒3 , 𝑒4 }
Payoff
1
0
4
-2
• Probability: Given
• Payoff: Given
• 𝐸=1
• 𝐸=
• 𝐸=
1
3
+0
1
6
1
+0+1+
3
5
= $.83
6
+4
1
−2
1
4
1
+ (−2)(4)
A player rolls a die and receives the # of $ = to
the # of dots on the die. What is the expected
value to play?
Roll
#1
#2
#3
#4
#5
#6
Probability
1/6
1/6
1/6
1/6
1/6
1/6
Payoff
$1
$2
$3
$4
$5
$6
1
1
1
1
1
1
𝐸=1
+2
+3
+4
+5
+6
6
6
6
6
6
6
21
𝐸=
= $3.50
6
If E = 0 then the “game” is fair
Same game – what must I change the
payoff if I roll a “1” to make the game fair?
0=𝑥
1
1
1
1
1
1
+2
+3
+4
+5
+6
6
6
6
6
6
6
1
10
0= 𝑥+
6
3
−10 1
= 𝑥
3
6
x = -20
A lab contains 10 microscopes, 2 are defective. If 4 are
chosen what is the Expected value of Defective?
Probabilities of 0,1, or 2 defectives:
𝐶 2,0 𝐶(8,4) 1
𝑝0 =
=
𝐶(10,4)
3
𝐶 2,1 𝐶(8,3)
8
𝑝1 =
=
𝐶(10,4)
15
𝐶 2,2 𝐶(8,2)
2
𝑝2 =
=
𝐶(10,4)
15
Assign payoffs of 0 (no defective)or 1, 2
since we are determining the expected #:
• 𝐸 = 0 ∙ 𝑝0 + 1 ∙ 𝑝1 + 2 ∙ 𝑝2 =
•𝐸 =0∙
1
3
+1∙
8
15
+2∙
2
15
=
4
5
• Talk about the answer.
• Can’t have 4/5 of a microscope?
• We can interpret this to mean that in the long run, we will
average “just under 1 defective microscope”
Expected Value of Bernoulli Trials:
• With “n” trials the expected # of successes is:
E=np
*Where “p” is the probability of successes on any single
trial.
MC Test contains 100 questions each w/ 4
choices. What is the expected # of correct
guesses?
• Answer: 25
• So using Bernoulli to explain:
1
𝐸 = 𝑛𝑝 = 100
= 25
4
HW WS: 8.3; #s 1-17odd,21, 25, 27