Answer :
Let's go through each question step by step with detailed solutions:
### Question 15:
Given [tex]\( P(E_1) = 0.41 \)[/tex], [tex]\( P(E_2) = 0.24 \)[/tex], and [tex]\( P(E_1 \cap E_2) = 0.2 \)[/tex], we need to find [tex]\( P(E_1 \cup E_2) \)[/tex].
From the formula for the union of two events:
[tex]\[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \][/tex]
Substitute the given values:
[tex]\[ P(E_1 \cup E_2) = 0.41 + 0.24 - 0.2 = 0.45 \][/tex]
Thus, the correct answer is:
B. 0.45
### Question 16:
We need to find [tex]\( P(E_2 \backslash E_1) \)[/tex].
The set difference [tex]\( E_2 \backslash E_1 \)[/tex] is the probability that event [tex]\( E_2 \)[/tex] occurs but not [tex]\( E_1 \)[/tex]. This can be found using:
[tex]\[ P(E_2 \backslash E_1) = P(E_2) - P(E_2 \cap E_1) \][/tex]
Substitute the given values:
[tex]\[ P(E_2 \backslash E_1) = 0.24 - 0.2 = 0.04 \][/tex]
Thus, the correct answer is:
B. 0.04
### Question 17:
A box contains 5 white, 6 red, and 4 black balls. If three balls are randomly taken out, what is the probability that the first ball is white and both the second and third balls are red?
First, calculate the total number of balls:
[tex]\[ 5 + 6 + 4 = 15 \][/tex]
Next, find the probability of the sequence (first white, second red, third red):
- Probability first ball is white [tex]\( P(\text{first white}) = \frac{5}{15} \)[/tex]
- After one white ball is taken, the total number of balls is 14, so the probability that the second ball is red [tex]\( P(\text{second red}) = \frac{6}{14} \)[/tex]
- After taking one red ball, there are 13 balls left, so the probability that the third ball is red [tex]\( P(\text{third red}) = \frac{5}{13} \)[/tex]
Putting it all together:
[tex]\[ P(\text{first white, second red, third red}) = \frac{5}{15} \times \frac{6}{14} \times \frac{5}{13} = 0.054945054945054944 \][/tex]
Simplifying the fraction yields:
[tex]\[ \frac{5 \times 6 \times 5}{15 \times 14 \times 13} = \frac{150}{2730} = \frac{5}{91} \][/tex]
Thus, the correct answer is:
D. [tex]\(\frac{5}{91}\)[/tex]
### Question 18:
Items produced by a certain company are subjected to two kinds of defects [tex]\( D_1 \)[/tex] and [tex]\( D_2 \)[/tex]. One out of eight items has neither defect.
Given this, the probability of an item having neither defect is:
[tex]\[ \frac{1}{8} = 0.125 \][/tex]
Thus, the correct answer is:
A. 0.125
### Question 19:
Odds against a certain event are [tex]\( 4:9 \)[/tex]. We need to find the probability of the non-occurrence of this event.
Given odds against an event being [tex]\( 4:9 \)[/tex], it means for every 13 events (sum of 4 and 9), 4 are non-occurrences.
The probability of non-occurrence is:
[tex]\[ \frac{4}{4+9} = \frac{4}{13} \][/tex]
Thus, the correct answer is:
B. [tex]\(\frac{4}{13}\)[/tex]
### Question 20:
A bag contains 20 marbles of which 12 are red, and the rest are blue and white. If the probability of obtaining a blue marble is 0.25, find the probability of obtaining a white marble.
Given:
- Total marbles = 20
- Red marbles = 12
- Probability of blue marble [tex]\( P(\text{blue}) = 0.25 \)[/tex]
- Number of blue marbles = [tex]\( 0.25 \times 20 = 5 \)[/tex]
The remaining marbles are white:
[tex]\[ 20 - 12 - 5 = 3 \][/tex]
Thus, the correct answer is:
B. [tex]\(\frac{3}{20}\)[/tex]
### Question 15:
Given [tex]\( P(E_1) = 0.41 \)[/tex], [tex]\( P(E_2) = 0.24 \)[/tex], and [tex]\( P(E_1 \cap E_2) = 0.2 \)[/tex], we need to find [tex]\( P(E_1 \cup E_2) \)[/tex].
From the formula for the union of two events:
[tex]\[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \][/tex]
Substitute the given values:
[tex]\[ P(E_1 \cup E_2) = 0.41 + 0.24 - 0.2 = 0.45 \][/tex]
Thus, the correct answer is:
B. 0.45
### Question 16:
We need to find [tex]\( P(E_2 \backslash E_1) \)[/tex].
The set difference [tex]\( E_2 \backslash E_1 \)[/tex] is the probability that event [tex]\( E_2 \)[/tex] occurs but not [tex]\( E_1 \)[/tex]. This can be found using:
[tex]\[ P(E_2 \backslash E_1) = P(E_2) - P(E_2 \cap E_1) \][/tex]
Substitute the given values:
[tex]\[ P(E_2 \backslash E_1) = 0.24 - 0.2 = 0.04 \][/tex]
Thus, the correct answer is:
B. 0.04
### Question 17:
A box contains 5 white, 6 red, and 4 black balls. If three balls are randomly taken out, what is the probability that the first ball is white and both the second and third balls are red?
First, calculate the total number of balls:
[tex]\[ 5 + 6 + 4 = 15 \][/tex]
Next, find the probability of the sequence (first white, second red, third red):
- Probability first ball is white [tex]\( P(\text{first white}) = \frac{5}{15} \)[/tex]
- After one white ball is taken, the total number of balls is 14, so the probability that the second ball is red [tex]\( P(\text{second red}) = \frac{6}{14} \)[/tex]
- After taking one red ball, there are 13 balls left, so the probability that the third ball is red [tex]\( P(\text{third red}) = \frac{5}{13} \)[/tex]
Putting it all together:
[tex]\[ P(\text{first white, second red, third red}) = \frac{5}{15} \times \frac{6}{14} \times \frac{5}{13} = 0.054945054945054944 \][/tex]
Simplifying the fraction yields:
[tex]\[ \frac{5 \times 6 \times 5}{15 \times 14 \times 13} = \frac{150}{2730} = \frac{5}{91} \][/tex]
Thus, the correct answer is:
D. [tex]\(\frac{5}{91}\)[/tex]
### Question 18:
Items produced by a certain company are subjected to two kinds of defects [tex]\( D_1 \)[/tex] and [tex]\( D_2 \)[/tex]. One out of eight items has neither defect.
Given this, the probability of an item having neither defect is:
[tex]\[ \frac{1}{8} = 0.125 \][/tex]
Thus, the correct answer is:
A. 0.125
### Question 19:
Odds against a certain event are [tex]\( 4:9 \)[/tex]. We need to find the probability of the non-occurrence of this event.
Given odds against an event being [tex]\( 4:9 \)[/tex], it means for every 13 events (sum of 4 and 9), 4 are non-occurrences.
The probability of non-occurrence is:
[tex]\[ \frac{4}{4+9} = \frac{4}{13} \][/tex]
Thus, the correct answer is:
B. [tex]\(\frac{4}{13}\)[/tex]
### Question 20:
A bag contains 20 marbles of which 12 are red, and the rest are blue and white. If the probability of obtaining a blue marble is 0.25, find the probability of obtaining a white marble.
Given:
- Total marbles = 20
- Red marbles = 12
- Probability of blue marble [tex]\( P(\text{blue}) = 0.25 \)[/tex]
- Number of blue marbles = [tex]\( 0.25 \times 20 = 5 \)[/tex]
The remaining marbles are white:
[tex]\[ 20 - 12 - 5 = 3 \][/tex]
Thus, the correct answer is:
B. [tex]\(\frac{3}{20}\)[/tex]
