Explanation
Let it be the events:
• A: A teenager texts while driving.
,
• B: A teenager has a car accident in the first year of driving.
From the word problem, we have:
[tex]\begin{gathered} P(A)=63\%=\frac{63}{100}=0.63 \\ P(B)=30\%=\frac{30}{100}=0.3 \\ P(A\cap B)=21\%=\frac{21}{100}=0.21 \end{gathered}[/tex]
The probability that a teenager gets into an accident in the first year of driving, given that he or she texts while driving is:
[tex]\begin{gathered} P(B|A)=\frac{P(B\cap A)}{P(A)} \\ P(B\lvert A)=\frac{P(A\operatorname{\cap}B)}{P(A)} \\ P(B\lvert A)=\frac{0.21}{0.63} \\ P(B\lvert A)=0.33 \end{gathered}[/tex]Answer
0.33
