Answer :
Let's solve the problem by determining the conditional probability that a senior takes the bus every day given that they have a driver's license.
Given probabilities:
- [tex]\( P(D) = 0.84 \)[/tex]: Probability that a senior has a driver's license.
- [tex]\( P(B) = 0.16 \)[/tex]: Probability that a senior takes the bus every day.
- [tex]\( P(D \cap B) = 0.14 \)[/tex]: Probability that a senior has a driver's license and takes the bus every day.
We need to find the conditional probability [tex]\( P(B|D) \)[/tex], which is the probability that a senior takes the bus every day given that they have a driver's license. The formula for conditional probability is:
[tex]\[ P(B|D) = \frac{P(D \cap B)}{P(D)} \][/tex]
Substituting the given values:
[tex]\[ P(B|D) = \frac{0.14}{0.84} \][/tex]
Dividing the two numbers:
[tex]\[ P(B|D) \approx 0.16666666666666669 \][/tex]
To express this probability as a percentage, we multiply by 100:
[tex]\[ P(B|D) \times 100 \approx 16.67\% \][/tex]
Rounding to the nearest whole percent, we get:
[tex]\[ 17\% \][/tex]
Therefore, the probability that a senior takes the bus to school every day, given that they have a driver's license, is approximately [tex]\( 17\% \)[/tex].
Answer: [tex]\( \boxed{17\%} \)[/tex]
Given probabilities:
- [tex]\( P(D) = 0.84 \)[/tex]: Probability that a senior has a driver's license.
- [tex]\( P(B) = 0.16 \)[/tex]: Probability that a senior takes the bus every day.
- [tex]\( P(D \cap B) = 0.14 \)[/tex]: Probability that a senior has a driver's license and takes the bus every day.
We need to find the conditional probability [tex]\( P(B|D) \)[/tex], which is the probability that a senior takes the bus every day given that they have a driver's license. The formula for conditional probability is:
[tex]\[ P(B|D) = \frac{P(D \cap B)}{P(D)} \][/tex]
Substituting the given values:
[tex]\[ P(B|D) = \frac{0.14}{0.84} \][/tex]
Dividing the two numbers:
[tex]\[ P(B|D) \approx 0.16666666666666669 \][/tex]
To express this probability as a percentage, we multiply by 100:
[tex]\[ P(B|D) \times 100 \approx 16.67\% \][/tex]
Rounding to the nearest whole percent, we get:
[tex]\[ 17\% \][/tex]
Therefore, the probability that a senior takes the bus to school every day, given that they have a driver's license, is approximately [tex]\( 17\% \)[/tex].
Answer: [tex]\( \boxed{17\%} \)[/tex]