Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
