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Zorluk: OrtaCompound Events and Probability Laws

In an archery competition, two archers, Kemi and Chidi, attempt to hit a target independently. The probability that Kemi hits the target is 13\frac{1}{3} and the probability that Chidi hits the target is 25\frac{2}{5}. What is the probability that at least one of them hits the target?

  1. 35\frac{3}{5}Cevap
  2. B
    1115\frac{11}{15}
  3. C
    215\frac{2}{15}
  4. D
    1315\frac{13}{15}

Cevap

The probability that at least one archer hits the target is 35\frac{3}{5}.
For independent events, the probability of both events occurring is P(KC)=P(K)×P(C)=13×25=215P(K \cap C) = P(K) \times P(C) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}. Applying the addition law of probability P(KC)=P(K)+P(C)P(KC)P(K \cup C) = P(K) + P(C) - P(K \cap C) gives 13+25215=915=35\frac{1}{3} + \frac{2}{5} - \frac{2}{15} = \frac{9}{15} = \frac{3}{5}.

Adım Adım Çözüm

1
Identify event probabilities and independence
P(Kemi)=13P(\text{Kemi}) = \frac{1}{3} and P(Chidi)=25P(\text{Chidi}) = \frac{2}{5}
The problem specifies that the two attempts are independent events.
2
Calculate the joint probability (intersection) using the multiplication law for independent events
P(KemiChidi)=P(Kemi)×P(Chidi)=13×25=215P(\text{Kemi} \cap \text{Chidi}) = P(\text{Kemi}) \times P(\text{Chidi}) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}
For independent events, the probability of both occurring together is the product of their individual probabilities.
3
Apply the addition law of probability to find the union
P(KemiChidi)=13+25215=5+6215=915=35P(\text{Kemi} \cup \text{Chidi}) = \frac{1}{3} + \frac{2}{5} - \frac{2}{15} = \frac{5 + 6 - 2}{15} = \frac{9}{15} = \frac{3}{5}
The addition law states P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B) to prevent double counting the intersection.

Anahtar Kavram

Compound Events, Multiplication Law for Independent Events, and Addition Law of Probability
Tahmini Süre:1m 30s
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