Certainly! Let's go through each part of the question step-by-step:
### Given Probabilities:
- Probability that Boy 1 passes: [tex]\(\frac{2}{3}\)[/tex]
- Probability that Boy 2 passes: [tex]\(\frac{5}{8}\)[/tex]
- Probability that Boy 3 passes: [tex]\(\frac{3}{4}\)[/tex]
### (i) Probability that all three boys passed:
To find the probability that all three boys pass, we multiply their individual probabilities:
[tex]\[
\text{P(All pass)} = \left(\frac{2}{3}\right) \times \left(\frac{5}{8}\right) \times \left(\frac{3}{4}\right) = 0.3125
\][/tex]
### (ii) Probability that none of the boys passed:
First, we need to find the probability that each boy fails:
- Probability that Boy 1 fails: [tex]\(1 - \frac{2}{3} = \frac{1}{3}\)[/tex]
- Probability that Boy 2 fails: [tex]\(1 - \frac{5}{8} = \frac{3}{8}\)[/tex]
- Probability that Boy 3 fails: [tex]\(1 - \frac{3}{4} = \frac{1}{4}\)[/tex]
Then we multiply these probabilities to find the probability that none of them pass:
[tex]\[
\text{P(None pass)} = \left(\frac{1}{3}\right) \times \left(\frac{3}{8}\right) \times \left(\frac{1}{4}\right) = 0.03125
\][/tex]
### (iii) Probability that only two of the boys passed:
To find the probability that exactly two boys pass and one fails, we consider all possible combinations where this could happen:
1. Boy 1 and Boy 2 pass, and Boy 3 fails:
[tex]\[
\left(\frac{2}{3}\right) \times \left(\frac{5}{8}\right) \times \left(\frac{1}{4}\right)
\][/tex]
2. Boy 1 and Boy 3 pass, and Boy 2 fails:
[tex]\[
\left(\frac{2}{3}\right) \times \left(\frac{1}{8}\right) \times \left(\frac{3}{4}\right)
\][/tex]
3. Boy 2 and Boy 3 pass, and Boy 1 fails:
[tex]\[
\left(\frac{1}{3}\right) \times \left(\frac{5}{8}\right) \times \left(\frac{3}{4}\right)
\][/tex]
Now, we add these probabilities together to find the overall probability that exactly two boys pass and one fails:
[tex]\[
\text{P(Only two pass)} = \left(\frac{2}{3} \times \frac{5}{8} \times \frac{1}{4}\right) + \left(\frac{2}{3} \times \frac{1}{8} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{5}{8} \times \frac{3}{4}\right) = 0.44791666666666663
\][/tex]
### Summary
To summarize, the probabilities are:
- (i) All three boys pass: [tex]\(0.3125\)[/tex]
- (ii) None of the boys pass: [tex]\(0.03125\)[/tex]
- (iii) Only two of the boys pass: [tex]\(0.44791666666666663\)[/tex]
