Answer :
Sure, let's work through this problem step by step.
a) Completing the Table
First, we complete the given table with the provided probabilities. The table initially looks like this:
| | Good shot | Bad shot |
|---------------|-----------|----------|
| Right club | 2/3 | ? |
| Wrong club | ? | 3/4 |
We need to fill in the missing probabilities.
- For the right club:
- The probability of a good shot with the right club is [tex]\( \frac{2}{3} \)[/tex].
- Therefore, the probability of a bad shot with the right club is [tex]\( 1 - \frac{2}{3} = \frac{1}{3} \)[/tex].
- For the wrong club:
- The probability of a bad shot with the wrong club is [tex]\( \frac{3}{4} \)[/tex].
- Therefore, the probability of a good shot with the wrong club is [tex]\( 1 - \frac{3}{4} = \frac{1}{4} \)[/tex].
Now the complete table looks like this:
| | Good shot | Bad shot |
|---------------|--------------|-------------|
| Right club | [tex]\( \frac{2}{3} \)[/tex] | [tex]\( \frac{1}{3} \)[/tex] |
| Wrong club | [tex]\( \frac{1}{4} \)[/tex] | [tex]\( \frac{3}{4} \)[/tex] |
Also, since she chooses the clubs randomly, the probability of choosing the right club among 5 clubs is [tex]\( \frac{1}{5} \)[/tex], and the probability of choosing the wrong club is [tex]\( 1 - \frac{1}{5} = \frac{4}{5} \)[/tex].
b) Probability of Reaching the Green in At Most Two Shots
We will now use a tree diagram approach to solve this.
1. First Shot:
- She could choose the right or wrong club.
- For the right club:
- Probability of a good shot [tex]\( = \frac{1}{5} \times \frac{2}{3} = \frac{2}{15} \)[/tex]
- Probability of a bad shot [tex]\( = \frac{1}{5} \times \frac{1}{3} = \frac{1}{15} \)[/tex]
- For the wrong club:
- Probability of a good shot [tex]\( = \frac{4}{5} \times \frac{1}{4} = \frac{1}{5} \)[/tex]
- Probability of a bad shot [tex]\( = \frac{4}{5} \times \frac{3}{4} = \frac{3}{5} \)[/tex]
The probability that she reaches the green in one good shot (first shot):
[tex]\[ P(\text{green in one shot}) = \frac{2}{15} + \frac{1}{5} = \frac{2}{15} + \frac{3}{15} = \frac{5}{15} = \frac{1}{3} \][/tex]
2. Second Shot (if the first shot is bad):
- For the right club chosen initially:
- Probability of a bad shot followed by a good shot:
[tex]\[ \frac{1}{5} \times \frac{1}{3} \times \frac{2}{3} = \frac{1}{5} \times \frac{2}{9} = \frac{2}{45} \][/tex]
- For the wrong club chosen initially:
- Probability of a bad shot followed by a good shot:
[tex]\[ \frac{4}{5} \times \frac{3}{4} \times \frac{1}{4} = \frac{4}{5} \times \frac{3}{16} = \frac{3}{20} \][/tex]
The probability that she reaches the green in two shots:
[tex]\[ P(\text{green in two shots}) = \frac{2}{45} + \frac{3}{20} \][/tex]
Convert [tex]\(\frac{3}{20}\)[/tex] to a common denominator with [tex]\( \frac{2}{45} \)[/tex]:
[tex]\[ \frac{3}{20} = \frac{27}{180}, \quad \frac{2}{45} = \frac{8}{180} \][/tex]
[tex]\[ \frac{8}{180} + \frac{27}{180} = \frac{35}{180} = \frac{7}{36} \][/tex]
Now, combining the probabilities from one and two shots:
[tex]\[ P(\text{green in at most two shots}) = \frac{1}{3} + \frac{7}{36} = \frac{12}{36} + \frac{7}{36} = \frac{19}{36} \][/tex]
So, summarizing:
- [tex]\(P(\text{good shot with right club}) = \frac{2}{3}\)[/tex]
- [tex]\(P(\text{bad shot with right club}) = \frac{1}{3}\)[/tex]
- [tex]\(P(\text{good shot with wrong club}) = \frac{1}{4}\)[/tex]
- [tex]\(P(\text{bad shot with wrong club}) = \frac{3}{4}\)[/tex]
- [tex]\(P(\text{choosing right club}) = \frac{1}{5}\)[/tex]
- [tex]\(P(\text{choosing wrong club}) = \frac{4}{5}\)[/tex]
- [tex]\(P(\text{green in one shot}) = \frac{1}{3}\)[/tex]
- [tex]\(P(\text{green in two shots}) = \frac{7}{36}\)[/tex]
- Total probability of reaching the green in at most two shots:
[tex]\[ P(\text{at most two shots}) = \frac{19}{36} = 0.52778... \][/tex]
So the probability that she reaches the green in at most two shots is [tex]\( \approx 0.528 \)[/tex].
a) Completing the Table
First, we complete the given table with the provided probabilities. The table initially looks like this:
| | Good shot | Bad shot |
|---------------|-----------|----------|
| Right club | 2/3 | ? |
| Wrong club | ? | 3/4 |
We need to fill in the missing probabilities.
- For the right club:
- The probability of a good shot with the right club is [tex]\( \frac{2}{3} \)[/tex].
- Therefore, the probability of a bad shot with the right club is [tex]\( 1 - \frac{2}{3} = \frac{1}{3} \)[/tex].
- For the wrong club:
- The probability of a bad shot with the wrong club is [tex]\( \frac{3}{4} \)[/tex].
- Therefore, the probability of a good shot with the wrong club is [tex]\( 1 - \frac{3}{4} = \frac{1}{4} \)[/tex].
Now the complete table looks like this:
| | Good shot | Bad shot |
|---------------|--------------|-------------|
| Right club | [tex]\( \frac{2}{3} \)[/tex] | [tex]\( \frac{1}{3} \)[/tex] |
| Wrong club | [tex]\( \frac{1}{4} \)[/tex] | [tex]\( \frac{3}{4} \)[/tex] |
Also, since she chooses the clubs randomly, the probability of choosing the right club among 5 clubs is [tex]\( \frac{1}{5} \)[/tex], and the probability of choosing the wrong club is [tex]\( 1 - \frac{1}{5} = \frac{4}{5} \)[/tex].
b) Probability of Reaching the Green in At Most Two Shots
We will now use a tree diagram approach to solve this.
1. First Shot:
- She could choose the right or wrong club.
- For the right club:
- Probability of a good shot [tex]\( = \frac{1}{5} \times \frac{2}{3} = \frac{2}{15} \)[/tex]
- Probability of a bad shot [tex]\( = \frac{1}{5} \times \frac{1}{3} = \frac{1}{15} \)[/tex]
- For the wrong club:
- Probability of a good shot [tex]\( = \frac{4}{5} \times \frac{1}{4} = \frac{1}{5} \)[/tex]
- Probability of a bad shot [tex]\( = \frac{4}{5} \times \frac{3}{4} = \frac{3}{5} \)[/tex]
The probability that she reaches the green in one good shot (first shot):
[tex]\[ P(\text{green in one shot}) = \frac{2}{15} + \frac{1}{5} = \frac{2}{15} + \frac{3}{15} = \frac{5}{15} = \frac{1}{3} \][/tex]
2. Second Shot (if the first shot is bad):
- For the right club chosen initially:
- Probability of a bad shot followed by a good shot:
[tex]\[ \frac{1}{5} \times \frac{1}{3} \times \frac{2}{3} = \frac{1}{5} \times \frac{2}{9} = \frac{2}{45} \][/tex]
- For the wrong club chosen initially:
- Probability of a bad shot followed by a good shot:
[tex]\[ \frac{4}{5} \times \frac{3}{4} \times \frac{1}{4} = \frac{4}{5} \times \frac{3}{16} = \frac{3}{20} \][/tex]
The probability that she reaches the green in two shots:
[tex]\[ P(\text{green in two shots}) = \frac{2}{45} + \frac{3}{20} \][/tex]
Convert [tex]\(\frac{3}{20}\)[/tex] to a common denominator with [tex]\( \frac{2}{45} \)[/tex]:
[tex]\[ \frac{3}{20} = \frac{27}{180}, \quad \frac{2}{45} = \frac{8}{180} \][/tex]
[tex]\[ \frac{8}{180} + \frac{27}{180} = \frac{35}{180} = \frac{7}{36} \][/tex]
Now, combining the probabilities from one and two shots:
[tex]\[ P(\text{green in at most two shots}) = \frac{1}{3} + \frac{7}{36} = \frac{12}{36} + \frac{7}{36} = \frac{19}{36} \][/tex]
So, summarizing:
- [tex]\(P(\text{good shot with right club}) = \frac{2}{3}\)[/tex]
- [tex]\(P(\text{bad shot with right club}) = \frac{1}{3}\)[/tex]
- [tex]\(P(\text{good shot with wrong club}) = \frac{1}{4}\)[/tex]
- [tex]\(P(\text{bad shot with wrong club}) = \frac{3}{4}\)[/tex]
- [tex]\(P(\text{choosing right club}) = \frac{1}{5}\)[/tex]
- [tex]\(P(\text{choosing wrong club}) = \frac{4}{5}\)[/tex]
- [tex]\(P(\text{green in one shot}) = \frac{1}{3}\)[/tex]
- [tex]\(P(\text{green in two shots}) = \frac{7}{36}\)[/tex]
- Total probability of reaching the green in at most two shots:
[tex]\[ P(\text{at most two shots}) = \frac{19}{36} = 0.52778... \][/tex]
So the probability that she reaches the green in at most two shots is [tex]\( \approx 0.528 \)[/tex].
