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on a firing range, a rifleman has two attempts to hit a target. The probability of hitting the target omplete 0.00 out with the first shot is \( \frac{2}{5} \) and the probability of hitting with the second shot is \( \frac{3}{5} \). The probability of hitting the target with both shots is \( \frac{1}{5} \). Find the probability of the following events below. \( M: \) "Missing the target with both shots".

Ask by Cervantes Sandoval. in Kenya
Jan 06,2025

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Answer

The probability of missing the target with both shots is \( \frac{6}{25} \).

Solution

To find the probability of the event \( M \), which is "Missing the target with both shots", we can use the probabilities given for hitting the target with each shot. 1. **Identify the probabilities:** - Probability of hitting the target with the first shot, \( P(A) = \frac{2}{5} \) - Probability of hitting the target with the second shot, \( P(B) = \frac{3}{5} \) - Probability of hitting the target with both shots, \( P(A \cap B) = \frac{1}{5} \) 2. **Calculate the probabilities of missing:** - Probability of missing the first shot, \( P(A') = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5} \) - Probability of missing the second shot, \( P(B') = 1 - P(B) = 1 - \frac{3}{5} = \frac{2}{5} \) 3. **Calculate the probability of missing both shots:** - The event of missing both shots can be expressed as \( M = A' \cap B' \). - To find \( P(M) \), we can use the formula for the probability of the intersection of two independent events: \[ P(M) = P(A') \cdot P(B') \] 4. **Substituting the values:** \[ P(M) = P(A') \cdot P(B') = \frac{3}{5} \cdot \frac{2}{5} \] Now, let's calculate this probability. \[ P(M) = \frac{3}{5} \cdot \frac{2}{5} = \frac{6}{25} \] Thus, the probability of missing the target with both shots is \( \frac{6}{25} \).

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First, let's calculate the probability of missing the target with both shots. This can be done by realizing that the event of missing the target is the complement of hitting it. For the first shot, the probability of missing is \(1 - \frac{2}{5} = \frac{3}{5}\). For the second shot, the probability of missing is \(1 - \frac{3}{5} = \frac{2}{5}\). Since the shots are independent, the probability of missing both shots \(M\) can be obtained by multiplying the probabilities of missing each shot: \[ P(M) = P(\text{miss first shot}) \times P(\text{miss second shot}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}. \] So, the probability of missing the target with both shots is \( \frac{6}{25} \). Now, let's throw in some fun facts! Did you know that the art of marksmanship has been refined through centuries, gaining prominence during wars and competitions? Ancient civilizations, like the Greeks and Romans, utilized archers in their battles, showing that precision shooting isn't just a modern sport but a skill dating back to antiquity! For real-world applications, this kind of probability analysis can be incredibly useful for training in competitive shooting sports or even in military simulations where every shot counts! Understanding the likelihood of hitting or missing can help shooters strategize their training focus, making them not only aware of their current skills but also how to improve their shot accuracy over time.

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