# Probability of Independent Events “Without Replacement”

Independent Events
Two events are called independent, if the occurrence of the first event does not depend (not influenced) on the occurence of the second event. Event A is said independent on B if the occurrence of event A is not depended (not influenced) be whether occur or failed of event B. For example in throwing of two dice at the same time, the event of the emergence of first die eyes is not depended on the emergence of the second die eyes. The independent events are also called the statistically independent events or stochastically independent events.

If two events are independent, then the probability of the occurence of the two events is equal to the multiplication of the probabilities of the two events.

Let the probability of event A is P(A) and the probability of event B is P(B). If events A and B are independent then the probability of the occurence of events A and B is:

P(A∩B)=P(A)∙P(B)

Example 3: In tossing once of the 2 balance dices at the same time, find the probability of
a. The emerging of the first die eye is 3 and the second die eye is 5.
b. The emerging of the first die eye is less than 3 and the second die eye is 5.
The explanation:
a. Suppose A is the event of the emerging of the first die eye is 3, then P(A)=⅙.
Suppose B is the event of the emerging of the second die eye is 5 then P(B)=⅙.
P(A∩B)=P(A)∙P(B)=⅙∙⅙=1/36

b. Suppose D is the event of the emerging of the second die eye is less than 3, then members of D are {1,2} and P(D)=2/6.
Suppose E is the event of the emerging of the second die eye is less than 5, then members of E are {1,2,3,4} and P(E)=4/6.
P(D∩E)=P(D)∙P(E)=2/6∙4/6=8/36

Try to determine the problem above by using the following sample space which are tossing of two balance dice. Then compare the results with the results obtained above. Exercise Competency Test 3 of Probability part III
1. Two balance dice are tossed once together. Probability of getting 2 of first die and getting 4 of second die equals …
A. 1/36 B. 1/18 C. 1/12 D. 1/9 E. ⅙
correct: , the explanation:
P=PIPII=⅙∙⅙=1/36

2. Two balance dice are tossed once together. Probability of getting prime of first die and getting odd of second die equals …
A. ⅙ B. ⅕ C. ¼ D. ⅓ E. ½
correct: C, the explanation:
prime number less than or equal to the number of sides of a die contains {2,3,5}.
Then the odd numbers contains {1,3,5}.

P=3/6∙3/6=½∙½=¼

3. A coin is tossed twice in a row. Probability of the emerging image side at the first tossing and number side at the second tossing equals …
A. ½ B. ⅓ C. ¼ D. ⅕ E. ⅙
correct: C, the explanation:
A coin has two sides.

P=PIPII=½∙½=¼

4. From a set of playing cards (bridge card) two cards are taken one by one “without replacement. Probability of both cards taken are aces =⋯
A. 2/52 B. 3/52 C. 4/52 D. 12/2652 E. 16/2704
correct: D, the explanation:
Playing cards contains 52 cards and 4 aces among the cards.
Phrase “without replacement” at this restricted context means both the wanted elements quantity and the sample space are reduced by an element on the second same treatment. RELATED POSTs