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Sagot :
Alright, let's solve the problem step-by-step.
### Problem Statement
We are given the following probabilities:
1. Probability that Awush passes the promotion exam is [tex]\( P(A) = \frac{1}{2} \)[/tex].
2. Probability that Ebow passes the promotion exam is [tex]\( P(E) = \frac{3}{4} \)[/tex].
3. Probability that T. Jami passes the promotion exam is [tex]\( P(T) = x \)[/tex].
Moreover, the probability that all three teachers fail the interview is given as [tex]\( \frac{1}{12} \)[/tex].
### (A) Value of [tex]\( x \)[/tex]
To find [tex]\( x \)[/tex], we need to use the complementary probabilities of failure for each teacher. Let these be represented as follows:
- Probability that Awush fails the exam is [tex]\( P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \)[/tex].
- Probability that Ebow fails the exam is [tex]\( P(E') = 1 - P(E) = 1 - \frac{3}{4} = \frac{1}{4} \)[/tex].
- Probability that T. Jami fails the exam is [tex]\( P(T') = 1 - P(T) = 1 - x \)[/tex].
The combined probability that all three teachers fail is given by the product of their individual probabilities of failure:
[tex]\[ P(A' \cap E' \cap T') = P(A') \cdot P(E') \cdot P(T') \][/tex]
Given:
[tex]\[ \frac{1}{2} \cdot \frac{1}{4} \cdot (1 - x) = \frac{1}{12} \][/tex]
Now, solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{2} \cdot \frac{1}{4} \cdot (1 - x) = \frac{1}{12} \][/tex]
[tex]\[ \frac{1}{8} \cdot (1 - x) = \frac{1}{12} \][/tex]
[tex]\[ 1 - x = \frac{1}{12} \div \frac{1}{8} \][/tex]
[tex]\[ 1 - x = \frac{1}{12} \times \frac{8}{1} \][/tex]
[tex]\[ 1 - x = \frac{8}{12} \][/tex]
[tex]\[ 1 - x = \frac{2}{3} \][/tex]
[tex]\[ x = 1 - \frac{2}{3} \][/tex]
[tex]\[ x = \frac{1}{3} \][/tex]
Thus, the value of [tex]\( x \)[/tex] is:
[tex]\[ \boxed{ \frac{1}{3} } \][/tex]
### (ii) Probability that only one passes the Interviews
To find this probability, we need to consider the following three exclusive events:
1. Awush passes, and both Ebow and T. Jami fail.
2. Ebow passes, and both Awush and T. Jami fail.
3. T. Jami passes, and both Awush and Ebow fail.
Calculating each event:
1. Probability that Awush passes, and both Ebow and T. Jami fail:
[tex]\[ P(A \cap E' \cap T') = P(A) \cdot P(E') \cdot P(T') \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{4} \cdot \left( 1 - \frac{1}{3} \right) \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} \][/tex]
[tex]\[ = \frac{1}{12} \cdot \frac{2}{3} \][/tex]
[tex]\[ = \frac{2}{36} \][/tex]
[tex]\[ = \frac{1}{18} \][/tex]
2. Probability that Ebow passes, and both Awush and T. Jami fail:
[tex]\[ P(A' \cap E \cap T') = P(A') \cdot P(E) \cdot P(T') \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{3}{4} \cdot \left( 1 - \frac{1}{3} \right) \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{2}{3} \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{2} \][/tex]
[tex]\[ = \frac{1}{4} \][/tex]
3. Probability that T. Jami passes, and both Awush and Ebow fail:
[tex]\[ P(A' \cap E' \cap T) = P(A') \cdot P(E') \cdot P(T) \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{3} \][/tex]
[tex]\[ = \frac{1}{2} \cdot \][/tex]
[tex]\[ = \frac{1}{4} \][/tex]
[tex]\[ = \frac{1}{24} \][/tex]
Summing these probabilities:
[tex]\[ P(\text{only one passes}) = \frac{1}{3} + \frac{1}{8} + \frac{1}{24} \][/tex]
[tex]\[ = \frac{9}{24} \cdot \frac{3}{24} \[ = \boxed{0.375} \][/tex]
### Problem Statement
We are given the following probabilities:
1. Probability that Awush passes the promotion exam is [tex]\( P(A) = \frac{1}{2} \)[/tex].
2. Probability that Ebow passes the promotion exam is [tex]\( P(E) = \frac{3}{4} \)[/tex].
3. Probability that T. Jami passes the promotion exam is [tex]\( P(T) = x \)[/tex].
Moreover, the probability that all three teachers fail the interview is given as [tex]\( \frac{1}{12} \)[/tex].
### (A) Value of [tex]\( x \)[/tex]
To find [tex]\( x \)[/tex], we need to use the complementary probabilities of failure for each teacher. Let these be represented as follows:
- Probability that Awush fails the exam is [tex]\( P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \)[/tex].
- Probability that Ebow fails the exam is [tex]\( P(E') = 1 - P(E) = 1 - \frac{3}{4} = \frac{1}{4} \)[/tex].
- Probability that T. Jami fails the exam is [tex]\( P(T') = 1 - P(T) = 1 - x \)[/tex].
The combined probability that all three teachers fail is given by the product of their individual probabilities of failure:
[tex]\[ P(A' \cap E' \cap T') = P(A') \cdot P(E') \cdot P(T') \][/tex]
Given:
[tex]\[ \frac{1}{2} \cdot \frac{1}{4} \cdot (1 - x) = \frac{1}{12} \][/tex]
Now, solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{2} \cdot \frac{1}{4} \cdot (1 - x) = \frac{1}{12} \][/tex]
[tex]\[ \frac{1}{8} \cdot (1 - x) = \frac{1}{12} \][/tex]
[tex]\[ 1 - x = \frac{1}{12} \div \frac{1}{8} \][/tex]
[tex]\[ 1 - x = \frac{1}{12} \times \frac{8}{1} \][/tex]
[tex]\[ 1 - x = \frac{8}{12} \][/tex]
[tex]\[ 1 - x = \frac{2}{3} \][/tex]
[tex]\[ x = 1 - \frac{2}{3} \][/tex]
[tex]\[ x = \frac{1}{3} \][/tex]
Thus, the value of [tex]\( x \)[/tex] is:
[tex]\[ \boxed{ \frac{1}{3} } \][/tex]
### (ii) Probability that only one passes the Interviews
To find this probability, we need to consider the following three exclusive events:
1. Awush passes, and both Ebow and T. Jami fail.
2. Ebow passes, and both Awush and T. Jami fail.
3. T. Jami passes, and both Awush and Ebow fail.
Calculating each event:
1. Probability that Awush passes, and both Ebow and T. Jami fail:
[tex]\[ P(A \cap E' \cap T') = P(A) \cdot P(E') \cdot P(T') \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{4} \cdot \left( 1 - \frac{1}{3} \right) \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} \][/tex]
[tex]\[ = \frac{1}{12} \cdot \frac{2}{3} \][/tex]
[tex]\[ = \frac{2}{36} \][/tex]
[tex]\[ = \frac{1}{18} \][/tex]
2. Probability that Ebow passes, and both Awush and T. Jami fail:
[tex]\[ P(A' \cap E \cap T') = P(A') \cdot P(E) \cdot P(T') \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{3}{4} \cdot \left( 1 - \frac{1}{3} \right) \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{2}{3} \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{2} \][/tex]
[tex]\[ = \frac{1}{4} \][/tex]
3. Probability that T. Jami passes, and both Awush and Ebow fail:
[tex]\[ P(A' \cap E' \cap T) = P(A') \cdot P(E') \cdot P(T) \][/tex]
[tex]\[ = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{3} \][/tex]
[tex]\[ = \frac{1}{2} \cdot \][/tex]
[tex]\[ = \frac{1}{4} \][/tex]
[tex]\[ = \frac{1}{24} \][/tex]
Summing these probabilities:
[tex]\[ P(\text{only one passes}) = \frac{1}{3} + \frac{1}{8} + \frac{1}{24} \][/tex]
[tex]\[ = \frac{9}{24} \cdot \frac{3}{24} \[ = \boxed{0.375} \][/tex]
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