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Sagot :
To solve this problem, we need to find the probabilities associated with two boys passing an examination.
Let:
- [tex]\(P(A)\)[/tex] be the probability that the first boy passes the examination, which is [tex]\(\frac{2}{3}\)[/tex].
- [tex]\(P(B)\)[/tex] be the probability that the second boy passes the examination, which is [tex]\(\frac{5}{8}\)[/tex].
### (i) Probability that both boys pass the examination
To find the probability that both boys pass the examination, we calculate the product of the individual probabilities:
[tex]\[P(A \text{ and } B) = P(A) \times P(B)\][/tex]
Given:
[tex]\[P(A) = \frac{2}{3}\][/tex]
[tex]\[P(B) = \frac{5}{8}\][/tex]
So:
[tex]\[P(A \text{ and } B) = \left(\frac{2}{3}\right) \times \left(\frac{5}{8}\right) = \frac{2 \times 5}{3 \times 8} = \frac{10}{24} = \frac{5}{12}\][/tex]
Numerically, this evaluates to approximately [tex]\(0.4167\)[/tex].
Thus, the probability that both boys pass the examination is [tex]\(\frac{5}{12}\)[/tex] or approximately [tex]\(0.4167\)[/tex].
### (ii) Probability that only one boy passes the examination
To find the probability that only one boy passes the examination, we need to consider two separate cases:
1. The first boy passes and the second boy fails.
2. The first boy fails and the second boy passes.
#### Case 1: The first boy passes and the second boy fails
[tex]\[P(A \text{ and not } B) = P(A) \times (1 - P(B))\][/tex]
Given:
[tex]\[P(A) = \frac{2}{3}\][/tex]
[tex]\[1 - P(B) = 1 - \frac{5}{8} = \frac{3}{8}\][/tex]
So:
[tex]\[P(A \text{ and not } B) = \left(\frac{2}{3}\right) \times \left(\frac{3}{8}\right) = \frac{2 \times 3}{3 \times 8} = \frac{6}{24} = \frac{1}{4}\][/tex]
Numerically, this evaluates to [tex]\(0.25\)[/tex].
#### Case 2: The first boy fails and the second boy passes
[tex]\[P(\text{not } A \text{ and } B) = (1 - P(A)) \times P(B)\][/tex]
Given:
[tex]\[1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3}\][/tex]
[tex]\[P(B) = \frac{5}{8}\][/tex]
So:
[tex]\[P(\text{not } A \text{ and } B) = \left(\frac{1}{3}\right) \times \left(\frac{5}{8}\right) = \frac{1 \times 5}{3 \times 8} = \frac{5}{24}\][/tex]
Numerically, this evaluates to approximately [tex]\(0.2083\)[/tex].
To find the total probability that only one boy passes, we add these two probabilities:
[tex]\[P(\text{only one passes}) = P(A \text{ and not } B) + P(\text{not } A \text{ and } B)\][/tex]
Given:
[tex]\[P(A \text{ and not } B) = \frac{1}{4} = 0.25\][/tex]
[tex]\[P(\text{not } A \text{ and } B) = \frac{5}{24} \approx 0.2083\][/tex]
So:
[tex]\[P(\text{only one passes}) = 0.25 + 0.2083 = 0.4583\][/tex]
Therefore, the probability that only one boy passes the examination is approximately [tex]\(0.4583\)[/tex].
Let:
- [tex]\(P(A)\)[/tex] be the probability that the first boy passes the examination, which is [tex]\(\frac{2}{3}\)[/tex].
- [tex]\(P(B)\)[/tex] be the probability that the second boy passes the examination, which is [tex]\(\frac{5}{8}\)[/tex].
### (i) Probability that both boys pass the examination
To find the probability that both boys pass the examination, we calculate the product of the individual probabilities:
[tex]\[P(A \text{ and } B) = P(A) \times P(B)\][/tex]
Given:
[tex]\[P(A) = \frac{2}{3}\][/tex]
[tex]\[P(B) = \frac{5}{8}\][/tex]
So:
[tex]\[P(A \text{ and } B) = \left(\frac{2}{3}\right) \times \left(\frac{5}{8}\right) = \frac{2 \times 5}{3 \times 8} = \frac{10}{24} = \frac{5}{12}\][/tex]
Numerically, this evaluates to approximately [tex]\(0.4167\)[/tex].
Thus, the probability that both boys pass the examination is [tex]\(\frac{5}{12}\)[/tex] or approximately [tex]\(0.4167\)[/tex].
### (ii) Probability that only one boy passes the examination
To find the probability that only one boy passes the examination, we need to consider two separate cases:
1. The first boy passes and the second boy fails.
2. The first boy fails and the second boy passes.
#### Case 1: The first boy passes and the second boy fails
[tex]\[P(A \text{ and not } B) = P(A) \times (1 - P(B))\][/tex]
Given:
[tex]\[P(A) = \frac{2}{3}\][/tex]
[tex]\[1 - P(B) = 1 - \frac{5}{8} = \frac{3}{8}\][/tex]
So:
[tex]\[P(A \text{ and not } B) = \left(\frac{2}{3}\right) \times \left(\frac{3}{8}\right) = \frac{2 \times 3}{3 \times 8} = \frac{6}{24} = \frac{1}{4}\][/tex]
Numerically, this evaluates to [tex]\(0.25\)[/tex].
#### Case 2: The first boy fails and the second boy passes
[tex]\[P(\text{not } A \text{ and } B) = (1 - P(A)) \times P(B)\][/tex]
Given:
[tex]\[1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3}\][/tex]
[tex]\[P(B) = \frac{5}{8}\][/tex]
So:
[tex]\[P(\text{not } A \text{ and } B) = \left(\frac{1}{3}\right) \times \left(\frac{5}{8}\right) = \frac{1 \times 5}{3 \times 8} = \frac{5}{24}\][/tex]
Numerically, this evaluates to approximately [tex]\(0.2083\)[/tex].
To find the total probability that only one boy passes, we add these two probabilities:
[tex]\[P(\text{only one passes}) = P(A \text{ and not } B) + P(\text{not } A \text{ and } B)\][/tex]
Given:
[tex]\[P(A \text{ and not } B) = \frac{1}{4} = 0.25\][/tex]
[tex]\[P(\text{not } A \text{ and } B) = \frac{5}{24} \approx 0.2083\][/tex]
So:
[tex]\[P(\text{only one passes}) = 0.25 + 0.2083 = 0.4583\][/tex]
Therefore, the probability that only one boy passes the examination is approximately [tex]\(0.4583\)[/tex].
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