# Conditional Probability, Chance of Next Event Happening

Using Tree Diagram
In the taking of two cards in a row from a set of cards, probability of getting ace on the first taking and heart on the second taking, can be determined by using a tree diagram as shown below.

Suppose in the taking of a card, the taken card is returned, then the tree diagram from the probability of the taking is as follows: The probability of the taking of ace on the first taking and the heart on the second taking
=4/52∙13/52=1/52

Suppose A states event of the first taking is ace and B states event of the second taking is hearth, then:
P(A∩B)=1/52

Conditional Probability
If two events occur sequentially and both events are not mutually exclusive, but influence each others, then the event is called a conditional event. Probability of occurence of the event B which will occur if event A is known that have occured, is called conditional probability and it is denoted by
P(B|A)

(read: probability of event B will occur if event A has occured).
Suppose in the election of a certain chairman, there are 6 male candidates where 4 of them are still actively working and 2 candidates have retired, 3 female candidates where 1 of them is still actively working and 2 have retired. Suppose one person will be selected randomly to be the chairman. How to determine the probability of the selected is male with the condition that he is actively working.

Suppose A represents the event which the selected is still working, and B represents the selected is male. P(B|A) will be determined. Probability of election can be described by the following tree diagram. Consider that P(A)=5/9;P(B|A)=⅘;P(A∩B)=4/9.
There is a relation between them ie:
P(A∩B)=P(A)∙P(B|A) or 4/9=5/9∙⅘

And this is according to
P(A∩B)=P(A)∙P(B|A)

Or can written as If event B occurs after event A has occured, then probability event B is called conditional probability, is denoted by P(B|A), and formulated width Example 4:
Two dice in balance are thrown simultaneously. If the emerging of the first die are odd number, determine the probability of the emerging of the sum of numeral sides facing up is less than 5.
The explanation:
Suppose A is the event of the emerging of the first die’s eyes is odd number. The members of A are

A={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
P(A)=18/36

Let B is the event of the emerging of the sum of dice’s eye is less than 5. The members of B are
B={(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)}
A∩B={(1,1),(1,2),(1,3),(3,1)}→P(A∩B)=4/36

Probability of event of the emerging of the sum of numeral sides is less than 5 if the first die side appears odd number is Example 5:
Probability of a certain wife watches TV alone =0.7. Probability of the wife and her husband watch TV together =0.4. Determine the probability of her husband watches TV after the wife had watched TV previously.
The explanation:
Suppose T is an event of the wife watches TV alone, then P(T)=0.7.
Suppose M is an event of her husband watches TV alone, then P(T∩M)=0.4.
Probability of her husband watches TV after the wife had watched TV previously is Exercise Competency Test 4 of Probability Part III
1. Two balance dice will be thrown at the same time. The sum of numeral sides facing up is less than 4 and the first die appearance is desired a 1. The satisfied chance of the second die throwing is …
A. ⅓ B. ⅔ C. 1/12 D. 1/9 E. ⅙
Correct: A, the explanation:
The chance of first die getting one equals ⅙. The elements of the sum which is less than 4 by the following first die getting are {(1,1),(1,2)}. The number of the possible outcomes of two dice rolling is 6×6=36. The chance of the following sum is 2/36.
Hence the satisfied chance of the second die throwing is 2. Probability of a certain wife watches TV alone =0.75. Probability of her husband watches TV alone =0.65. Probability of the wife or hers watch TV =0.9. The probability that the wife joins watching TV when hers is watching TV =⋯
A. 1/15 B. 1/14 C. 10/13 D. 5/9 E. 5/8
Correct: C, the explanation:
The chance of each that person or both watching TV is 0.9 then the chance of both persons watching TV together is 0.75+0.65-0.9=0.5.
These recently statements can be shown as A is the wife, H is her husband and S is the sample space.
The probability that the wife joins her husband for watching TV equals
0.5/0.65=50/65=10/13

3. A company plans to select employees for training. There are 5 male candidates: 3 from the personnel division and 2 from EDP division, 3 female candidates: 1 from the personnel division and 2 of the EDP one. Probability of the selected to be training is a man with the condition is from the EDP division is equal to …
A. 0.1 B. 0.2 C. 0.3 D. 0.4 E. 0.5
Correct: D, the explanation:
The chance of a man selected with the condition is from the EDP is depended by number of men in EDP. Its chance is ⅖=0.4

4. Probability of a flight departs on time =0.80, the probability that the flight lands on time =0.90 and the flight departs and lands on time =0.792. If the flight is known to depart on time, then the probability of the flight lands on time =⋯
A. 0.77 B. 0.85 C. 0.88 D. 0.95 E. 0.99
Correct: E, the explanation: The probability of (B|A) can not be greater than 1.0. But there is a not-matching statement on being known that the probability of flight landing on time equals 0.90.

5. Probability correct diagnosis of a doctor about a disease =0.75. If known the diagnosis of a doctor is wrong, the probability that a patient will prosecute to the court =0.92. Probability that the diagnosis of a doctor is wrong and a patient prosecutes him, is equal to …
A. 0.96 B. 0.69 C. 0.32 D. 0.23 E. 0.135
Correct: B, the explanation:
The patient might not prosecute the doctor, if the doctor had not made a wrong diagnosis on him/her.

P(A)=0.75;P(B|A)=0.92
P(A∩B)=P(A)×P(B|A)
P(A∩B)=0.75×0.92=0.69

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