Probability Of Compound Events – Two or More Independent Events

The Range of Probability’s Value
Many examples and problemsthat show the value of probability of an event that are given above. If we look at the value of probabilities that are obtained they lie between 0 to 1.
Because the event is a subset of the sample space, then the number of members of event are less than or equal to the number of members of the sample space, or can be written: 0≤n(K)≤n(S).
Since the probability of event K is P(K)=n(K)/n(S) then 0≤P(K)≤1.
If the probability of an event is equal to 0, then the event is called an impossible event to happen or not possible to happen. For example human can life at the sun, this is an impossible event. If the probability of an event is equal to 1, then the event is called by events that surely will happen. For example the night will be dark if there is no light, this is an event that surely will happen.
If S is the sample space, then the probability of S is equal to 1. P(S)=1.
If K ̅ is an event which is not the event K then P(K ̅ )=1-P(K).
Example 5: Suppose a die is thrown. Determine the probability:
a. The emerging of die’s eyes =7
b. The emerging of die’s eyes is more than 4
Answer:
A die is thrown, then the sample space are:

S={1,2,3,4,5,6} and n(S)=6

a. Suppose K1 is the event of the emerging of die’s eyes =7
Because there is no die’s eyes which is numbered by 7 then n(K1)=0
Probability of the emerging of die’s eyes =7 is equal to 0 or P(K1)=0
b. Suppose K2 is the event of the emerging of die’s eyes is more than 4,
then K2={5,6} and n(K2)=2. The probability of the emerging of die’s eyes is more than 4 is P(K2)=2/6=⅓
Example 6: Suppose two dice are thrown together. Determine the probability of:
a. The sum of dice’s eyes that appear is less than 4.
b. The sum of dice’s eyes that appear is ≥4
Answer:
Two dice are thrown together then n(S)=62=36.
Suppose x is the sum of dice’s eyes that appear.
a. The event of the sum of dice’s eyes that appear is less than 4 are:

K={(1,1),(1,2),(2,1)}→n(K)=3

The probability of event K is P(K)=P(x<4)=3/36=1/12
b. The event of the sum of dice’s eyes that appear is ≥4 is K ̅. The probability of event K ̅ is:

Exercise Competency Test 5
1. Two dice are thrown together. The event below that has a 0 probability is …
A. The sum of dice’s eyes that appear =6
B. The multiplication of dice’s eyes that appear =10
C. The sum of dice’s eyes that appear is prime number
D. The multiplication of dice’s eyes that appear =12
E. The sum of dice’s eyes that appear >12
Answer: E, explanation:
Maximum value of a die’s side is 6 then the maximum value of the sum of two dice’s side is 12. Thus, The sum of dice’s eyes is unable to be upper than 12.

2. Probability of a pregnant woman gave birth to a baby boy or baby girl =….
A. 0,25 B. 0,5 C. 0,75 D. 0,90 E. 1
Answer: B, explanation:
It is as similar as a coin that has two sides.

P=½ =0,5
3. Suppose two coins are tossed at the same time, then the probability of the top side (readable) that appear are same =….
A. 1 B. 0,75 C. 0,5 D. 0,25 E. 0,1
Answer: C, explanation:
K states the expected event occuring. K={EE,NN}→n(K)=2.
The number of members of the sample space is n(K)=22=4

4. Suppose from a set of dominoes is drawn one card. The events below that will surely happen is …
A. The sum of card’s eyes <1 B. The sum of card’s eyes >1
C. The sum of card’s eyes <12 D. The sum of card’s eyes ≤12 E. The sum of card’s eyes >12
Answer: D, explanation:
Exactly, {0≤x≤12,xϵE}. P means enumeration number.

5. Suppose three coins are tossed together. Probability of the facing up (the readable side) is at least one image side =…
A. ⅛ B. ⅟9 C. ⅝ D. ⅞ E. 8/9
Answer: D, explanation:
We need to count probability of none image side (E) facing up first, it means all number sides (N) in face up position.


Number of the sample space is


The probability of at least one image side in face up position is

6. From a set of playing cards, one card is taken. Probability of that is drawn not Ace, is equal to …
A. 51/52 B. 48/52 C. 39/52 D. 36/52 E. 26/52
Answer: B, explanation:
We need to count probability of Ace cards drawing first.


Probability of not Ace cards drawing is

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