Probability Presentation 2

```Probability
Discussion Topics
 Conditional Probability
 Probability using the general multiplication rule
 Independent event
Question 1
The data in the table below represent the marital status of males and females
18 years old or older
Males
(In millions)
Females
(In millions)
Totals
(In millions)
Never married
30.3
25.0
55.3
Married
63.6
64.1
127.7
Widowed
2.6
11.3
13.9
Divorce
9.7
13.1
22.8
Totals
106.2
113.5
219.7
•Calculate the probability that a randomly selected individual 18 years or
older is widowed
•Calculate the probability that a randomly selected individual 18 years or
older is widowed and female
Solution
Probability that selected individual is widowed:
(a) Define W as the event of picking a widow
P(W)=Total of widows/Overall totals
F6/F8
Solution
Probability that selected individual is widowed and female
P(Widowed | Female)=Number of widowed/Number of females
E6/E8
Question 2
In the game of roulette, the wheel has slots numbered 0,00, and 1 through
36.A metal ball is allowed to roll around a wheel until it falls into one of the
numbered slots. You decide to play the game and place a bet on the number
17.
(a) What is the probability that the ball will land in the slot numbered 17 two
times in a row?
Solution
(1/38)/(1/38)
The events are independent because the probability of the 1st spin cannot affect the
probability of the second spin.
The ball does not remember it landed on 17 on the first spin
Question 3
Suppose that a box of 100 circuits is sent to a manufacturing plant. Of the 100
circuits shipped, 5 are defective. The plant manager receiving the circuits
randomly selects 2 and tests them. If both circuits work, she will accept the
shipment. Otherwise the shipment is rejected.
(a) Use a tree diagram to represent all possible outcomes
(b) What is the probability that the plant manager discovers at least 1
defective circuit and rejects the shipment
Tree diagram
Tool used is Microsoft Visio.
From our tree diagram and using addition rule:
P(at least 1 defective) = P(GD) + P(DG) + P(DD)
= 0.048+0.048+0.002
= 0.098
There is 0.098 probability of shipment being rejected
2nd Approach:
P(at least 1 defective) = 1-P(none defective)
= 1- (95/100)*(94/99)
= 1- 0.902
= 0.098
http://www.cis.famu.edu/~cdellor/math/
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