Keywords: Probability, target hitting, independent events.
Required Approach: Factual and analytical. This question requires calculating probabilities based on given data and applying probability rules.
Points to Remember:
- Probability of an event = (Favorable outcomes) / (Total outcomes)
- Independent events: The outcome of one event does not affect the outcome of another.
- Probability of two independent events both occurring: P(A and B) = P(A) * P(B)
- Probability of at least one of two events occurring: P(A or B) = P(A) + P(B) – P(A and B)
Introduction:
This problem involves calculating probabilities related to independent events. A, B, and C are three individuals attempting to hit a target, each with a different probability of success. We need to determine the probability of specific combinations of successes among them. We will use basic probability principles to solve this problem. The probabilities of each person hitting the target are: P(A) = 3/5, P(B) = 2/5, and P(C) = 3/4. These probabilities are assumed to be independent; the success or failure of one person does not influence the others.
Body:
(a) Probability of Any Two Hitting the Target:
This can happen in three ways: AB, AC, or BC. We will calculate the probability of each scenario and then sum them, ensuring we don’t double-count any overlapping scenarios (which is not the case here since the events are independent).
- P(A and B): P(A) * P(B) = (3/5) * (2/5) = 6/25
- P(A and C): P(A) * P(C) = (3/5) * (3/4) = 9/20
- P(B and C): P(B) * P(C) = (2/5) * (3/4) = 6/20 = 3/10
Therefore, the probability of any two hitting the target is: P(AB) + P(AC) + P(BC) = 6/25 + 9/20 + 3/10 = (24 + 45 + 30)/100 = 99/100
(b) Probability of At Least Two Hitting the Target:
This includes the scenarios where exactly two hit the target (calculated above) and the scenario where all three hit the target.
- P(A and B and C): P(A) * P(B) * P(C) = (3/5) * (2/5) * (3/4) = 18/100 = 9/50
Therefore, the probability of at least two hitting the target is: P(exactly two) + P(all three) = 99/100 + 9/50 = (99 + 18)/100 = 117/100. This result is greater than 1, which is impossible. There’s an error in the calculation of part (a). Let’s recalculate.
The correct calculation for (a) is:
P(exactly two) = P(A and B and not C) + P(A and not B and C) + P(not A and B and C)
= (3/5)(2/5)(1-3/4) + (3/5)(1-2/5)(3/4) + (1-3/5)(2/5)(3/4)
= (3/5)(2/5)(1/4) + (3/5)(3/5)(3/4) + (2/5)(2/5)(3/4)
= 6/100 + 27/100 + 12/100 = 45/100 = 9/20
Now, the probability of at least two hitting the target is:
P(at least two) = P(exactly two) + P(all three) = 9/20 + 9/50 = 45/100 + 18/100 = 63/100 = 0.63
Conclusion:
The probability of any two of A, B, and C hitting the target is 9/20 or 0.45. The probability of at least two of them hitting the target is 63/100 or 0.63. These calculations demonstrate the application of basic probability principles to independent events. Further analysis could involve exploring scenarios with varying probabilities of success for each individual or considering the impact of factors influencing accuracy, such as target distance or environmental conditions. A more complex model could incorporate these variables for a more realistic assessment. This analysis highlights the importance of understanding probability in various contexts, from simple games of chance to more complex real-world situations.
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