Competency Evaluation of Ability on Probability Teory Comprehension
Instruction: Choose the one correct answer of each question.
1. Two balance coins are thrown at the same time. Probability of emergence at least one numeral side …
A. ¼ B. ½ C. ¾ D. ⅔ E. ⅚
Correct: C, the explanation:
We don’t need to count each probability. Probabilities of one numeral and two numeral don’t need to be counted!
We need to count the probability of complement of that event first.
The sample space of throwing a coin once is equal to 2.
The sample space of throwing k coins once is equal to 2^k.
Thus the sample space of throwing 2 coins once is equal to n(S)=2²=4.
Suppose E states image side and N states numeral side. Suppose A consist of emergences at least one numeral side and Ᾱ is complement of event A. Thus Ᾱ consist of only emergence zero numeral sides. Thus

Ᾱ={EE}→n(Ᾱ)=1
P(Ᾱ)=¼
The probability of event A is equal to
P(A)=1-P(Ᾱ)
P(A)=1-¼=¾

2. There are 7 red marbles and 3 blue ones inside a box. From the box 3 marbles was taken once. Probability of at least one blue marble drawn =⋯
A. 7/24 B. 9/24 C. 11/24 D. 13/24 E. 17/24
Correct: E, the explanation:
Similar on the explanation question number one above.
We don’t need to count each probability. Probabilities of one, two and three blue marbles don’t need to be counted!
We need to count the probability of complement of that event first.
Suppose P(A) is the probability of at least one blue marble drawn and Ᾱ is complement of event A. Thus P(Ᾱ) is the probability of zero blue marbles and three ret ones drawn. Drawing marbles is a problem that can be solved with combination basic formula. Remember the basic formula of combination.

The probability of at least one blue marble drawn is equal to
P(A)=1-P(Ᾱ)=1-7/24=17/24

3. Two balance dice are thrown together. Suppose X represents the sum of dice eye emergences. P(6≤x≤8)=⋯
A. 1/9 B. 2/9 C. 4/9 D. ⅙ E. ⅓
Correct: C, the explanation:
The elements of X are {6,7,8} that can be written as

{(1+5),(2+4),(3+3),(4+2),(5+1),(1+6),(2+5),(3+4),(4+3),(5+2),(6+1),(2+6),(3+5),(4+4),(5+3),(6+2)}
n(X)=16
The sample space of throwing two balance dice equals n(S)=6×6=36. The chance P(6≤x≤8) equals

4. A bag contains 5 red balls and 3 white balls and 2 green ones. One ball is withdrawn, probability of the withdrawn ball is red or white =⋯
A. 0.5 B. 0.6 C. 0.7 D. 0.8 E. 0.9
Correct: D, the explanation:
Probability of event A and B happen is given by
P(A∩B)=P(A)×P(B)
Its meaning, occurrence of event A must be followed by event B.
Probability of event A or B happen is given by
P(A∪B)=P(A)+P(B)-P(A∩B)
At this moment, Each ball won’t have two colors. Hence
P(A∩B)=0
P(A∪B)=P(A)+P(B)
The probability of drawing red ball or white one equals
P(A∪B)=5/10+3/10=0.8

5. If events A and B are independent then the probability of occurence of event A with the condition that event B has occured =⋯
A. P(A∪B) D. P(A)+P(B)-P(A∩B)
B. P(A∩B) E. P(A∪B)-P(A∩B)
C. P(A)
Correct: C, the explanation:
Clear. If A depends on B then Its probability is written as P(A|B).

6. Two balance dice, each die has 4 red sides and 2 white sides. If those two dice are thrown once, then the probability of getting white sides is equal to …
A. 1/36 B. 1/9 C. ⅙ D. 4/9 E. 8/9
Correct: B, the explanation:

7. From a set of bridge card, two cards are taken one by one with replacement. Probability of getting heart first and spade at the second =⋯
A. 1/36 B. 1/169 C. 1/9 D. ½ E. ¾
Correct: B, the explanation:
A set of bridge card contains 52 cards. It consist of 4 types those are red diamond, black spade, red heart, black club.

Each type has plenty of cards as many as other.
Hence the chances of drawing a card of respective types with replacement are equal.
P=4/52=1/13
The probability of getting heart first and spade at the second is equal to
P=1/13×1/13=1/169

8. A bag contains 8 red marbles and 13 white marbles. Two marbles are drawn one by one without replacement. The probability of both marbles that drawn are red =⋯
A. 1/15 B. 2/15 C. ⅕ D. 4/15 E. ⅓
Correct: B, the explanation:
Phrase “without replacements” means that both the number of expected elements and the number of whole elements are reduced by one after withdrawing.

9. In a city there are 900 graduates, of them there are 460 employed men and 40 are unemployed , and 140 are employed women and 260 are unemployed. Suppose one of them is elected to be a certain chairman. If a male is chosen, then the probability of the male chosen is unemployed equals …
A. 23/30 B. 2/23 C. 2/25 D. 22/30 E. 4/25
Correct: C, the explanation:
Remember term Conditional Probability.
The number of men is 40+460=500
P=40/500=2/25

10. There are 4 red balls and 3 white ones inside first box. There are 7 red balls and 2 white ones inside second box. One ball is taken from each box randomly. The probability of a white ball taken from the first box and a red ball taken from the second box is equal to …
A. 28/63 B. 21/63 C. 8/63 D. 6/63 E. 5/63
Correct: B, the explanation: