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The probability of a man hitting a target is 1/10. The least number of shots required, so that the probability of his hitting the target at least once is greater than 1/4, is ______ (in integer).
Correct answer is '3'. Can you explain this answer?
Verified Answer
The probability of a man hitting a target is 1/10.The least number of ...
We have,
1 – (Probability of all shots result in failure) > 1/4
Using binomial distribution,
So the least number of shots required is 3.
1 – (Probability of all shots result in failure) > 1/4
Using binomial distribution,
So the least number of shots required is 3.
Most Upvoted Answer
The probability of a man hitting a target is 1/10.The least number of ...
To determine the least number of shots required for a man to hit a target with a probability greater than 1/4, we need to analyze the probabilities involved.
Understanding the Probability of Hitting the Target
- The probability of hitting the target in a single shot (p) = 1/10.
- The probability of missing the target in a single shot (q) = 1 - p = 1 - 1/10 = 9/10.
Calculating Probability of Hitting at Least Once
- If the man takes 'n' shots, the probability of missing the target in all n shots is given by:
\[
P(\text{missing all n shots}) = q^n = \left(\frac{9}{10}\right)^n
\]
- Therefore, the probability of hitting the target at least once is:
\[
P(\text{hitting at least once}) = 1 - P(\text{missing all n shots}) = 1 - \left(\frac{9}{10}\right)^n
\]
Setting Up the Inequality
- We want this probability to be greater than 1/4:
\[
1 - \left(\frac{9}{10}\right)^n > \frac{1}{4}
\]
- Rearranging gives:
\[
\left(\frac{9}{10}\right)^n < />
\]
Solving the Inequality
- Taking logarithms on both sides:
\[
n \cdot \log\left(\frac{9}{10}\right) < />
\]
- Since \(\log\left(\frac{9}{10}\right)\) is negative, we invert the inequality:
\[
n > \frac{\log\left(\frac{3}{4}\right)}{\log\left(\frac{9}{10}\right)}
\]
- Calculating the right side gives approximately:
\[
n > 2.5
\]
- The smallest integer satisfying this is \(n = 3\).
Conclusion
- Thus, the least number of shots required for the probability of hitting the target at least once to be greater than 1/4 is 3.
Understanding the Probability of Hitting the Target
- The probability of hitting the target in a single shot (p) = 1/10.
- The probability of missing the target in a single shot (q) = 1 - p = 1 - 1/10 = 9/10.
Calculating Probability of Hitting at Least Once
- If the man takes 'n' shots, the probability of missing the target in all n shots is given by:
\[
P(\text{missing all n shots}) = q^n = \left(\frac{9}{10}\right)^n
\]
- Therefore, the probability of hitting the target at least once is:
\[
P(\text{hitting at least once}) = 1 - P(\text{missing all n shots}) = 1 - \left(\frac{9}{10}\right)^n
\]
Setting Up the Inequality
- We want this probability to be greater than 1/4:
\[
1 - \left(\frac{9}{10}\right)^n > \frac{1}{4}
\]
- Rearranging gives:
\[
\left(\frac{9}{10}\right)^n < />
\]
Solving the Inequality
- Taking logarithms on both sides:
\[
n \cdot \log\left(\frac{9}{10}\right) < />
\]
- Since \(\log\left(\frac{9}{10}\right)\) is negative, we invert the inequality:
\[
n > \frac{\log\left(\frac{3}{4}\right)}{\log\left(\frac{9}{10}\right)}
\]
- Calculating the right side gives approximately:
\[
n > 2.5
\]
- The smallest integer satisfying this is \(n = 3\).
Conclusion
- Thus, the least number of shots required for the probability of hitting the target at least once to be greater than 1/4 is 3.
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The probability of a man hitting a target is 1/10.The least number of shots required, so that the probability of his hitting the target at least once is greater than 1/4,is ______ (in integer).Correct answer is '3'. Can you explain this answer?
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