Answer :
To determine the probability that the bus arrives either early or on time, we need to add the probabilities of each event occurring. Here are the detailed steps for solving this:
1. Identify the probabilities:
- The probability that the bus arrives early at the bus stop is given as [tex]\(\frac{4}{7}\)[/tex].
- The probability that the bus arrives on time is given as [tex]\(\frac{3}{14}\)[/tex].
2. Find a common denominator:
- To add these fractions, we need to have a common denominator.
In this case, the denominators are 7 and 14. The least common multiple of 7 and 14 is 14. Therefore, we convert [tex]\(\frac{4}{7}\)[/tex] to a fraction with a denominator of 14.
3. Convert [tex]\(\frac{4}{7}\)[/tex] to a fraction with a denominator of 14:
[tex]\[ \frac{4}{7} = \frac{4 \times 2}{7 \times 2} = \frac{8}{14} \][/tex]
4. Add the two probabilities:
- Now, we add [tex]\(\frac{8}{14}\)[/tex] (the converted probability of arriving early) and [tex]\(\frac{3}{14}\)[/tex] (the probability of arriving on time):
[tex]\[ \frac{8}{14} + \frac{3}{14} = \frac{8 + 3}{14} = \frac{11}{14} \][/tex]
5. Simplify the result:
- The fraction [tex]\(\frac{11}{14}\)[/tex] is already in its simplest form as 11 and 14 have no common factors other than 1.
Thus, the probability that the bus arrives early or on time is [tex]\(\frac{11}{14}\)[/tex].
1. Identify the probabilities:
- The probability that the bus arrives early at the bus stop is given as [tex]\(\frac{4}{7}\)[/tex].
- The probability that the bus arrives on time is given as [tex]\(\frac{3}{14}\)[/tex].
2. Find a common denominator:
- To add these fractions, we need to have a common denominator.
In this case, the denominators are 7 and 14. The least common multiple of 7 and 14 is 14. Therefore, we convert [tex]\(\frac{4}{7}\)[/tex] to a fraction with a denominator of 14.
3. Convert [tex]\(\frac{4}{7}\)[/tex] to a fraction with a denominator of 14:
[tex]\[ \frac{4}{7} = \frac{4 \times 2}{7 \times 2} = \frac{8}{14} \][/tex]
4. Add the two probabilities:
- Now, we add [tex]\(\frac{8}{14}\)[/tex] (the converted probability of arriving early) and [tex]\(\frac{3}{14}\)[/tex] (the probability of arriving on time):
[tex]\[ \frac{8}{14} + \frac{3}{14} = \frac{8 + 3}{14} = \frac{11}{14} \][/tex]
5. Simplify the result:
- The fraction [tex]\(\frac{11}{14}\)[/tex] is already in its simplest form as 11 and 14 have no common factors other than 1.
Thus, the probability that the bus arrives early or on time is [tex]\(\frac{11}{14}\)[/tex].
