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A skating rink attendant monitored the number of injuries at the rink over the past year. He tracked the ages of those injured and the kinds of skates worn during injury.\newlineThe probability that an injured skater was a minor is 0.80.8, the probability that an injured skater was wearing in-line skates is 0.30.3, and the probability that an injured skater was a minor and was wearing in-line skates is 0.20.2.\newlineWhat is the probability that a randomly chosen injured skater was a minor or was wearing in-line skates?\newlineWrite your answer as a whole number, decimal, or simplified fraction.

Full solution

Q. A skating rink attendant monitored the number of injuries at the rink over the past year. He tracked the ages of those injured and the kinds of skates worn during injury.\newlineThe probability that an injured skater was a minor is 0.80.8, the probability that an injured skater was wearing in-line skates is 0.30.3, and the probability that an injured skater was a minor and was wearing in-line skates is 0.20.2.\newlineWhat is the probability that a randomly chosen injured skater was a minor or was wearing in-line skates?\newlineWrite your answer as a whole number, decimal, or simplified fraction.
  1. Define Events: Let's call the events:\newlineC: The skater was a minor.\newlineD: The skater was wearing in-line skates.\newlineWe know:\newlineP(C)=0.8P(C) = 0.8\newlineP(D)=0.3P(D) = 0.3\newlineP(C and D)=0.2P(C \text{ and } D) = 0.2
  2. Use Addition Rule: We gotta find P(C or D)P(C \text{ or } D). So we use the addition rule:\newlineP(C or D)=P(C)+P(D)P(C and D)P(C \text{ or } D) = P(C) + P(D) - P(C \text{ and } D)
  3. Calculate P(C or D)P(C \text{ or } D): Now we do the math:\newlineP(C or D)=0.8+0.30.2P(C \text{ or } D) = 0.8 + 0.3 - 0.2
  4. Final Result: So that gives us:\newlineP(C or D)=1.10.2P(C \text{ or } D) = 1.1 - 0.2

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