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A survey of the 12th-grade students at Gaffigan High School found that [tex]$84\%$[/tex] of the seniors have their driver's licenses, [tex]$16\%$[/tex] of seniors take the bus every day to school, and [tex]$14\%$[/tex] of the seniors have driver's licenses and take the bus to school every day.

To the nearest whole percent, what is the probability that a senior takes the bus to school every day, given that he or she has a driver's license?

A. [tex]$17\%$[/tex]
B. [tex]$19\%$[/tex]
C. [tex]$36\%$[/tex]
D. [tex]$70\%$[/tex]



Answer :

GinnyAnswer
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]

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