Question:
Two persons \(A\) and \(B\) take part in a shooting competition. \(A\) can hit the target with probability \(0.6\), and \(B\) can hit the target with probability \(0.8\). \(A\) has the first shot, after which they strike alternately. Then, the probability that \(A\) wins the competition is:
Two persons \(A\) and \(B\) take part in a shooting competition. \(A\) can hit the target with probability \(0.6\), and \(B\) can hit the target with probability \(0.8\). \(A\) has the first shot, after which they strike alternately. Then, the probability that \(A\) wins the competition is:
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In alternate shooting problems, write all winning cases as an infinite geometric series. The common ratio is usually the probability that both players miss in one full round.
Updated On: Jun 18, 2026
- \(\frac{7}{10}\)
- \(\frac{15}{23}\)
- \(\frac{2}{3}\)
- \(\frac{11}{17}\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the probabilities of hitting and missing.
Probability that \(A\) hits the target is \[ P(A)=0.6=\frac{3}{5}. \] Therefore, probability that \(A\) misses is \[ 1-\frac{3}{5}=\frac{2}{5}. \] Probability that \(B\) hits the target is \[ P(B)=0.8=\frac{4}{5}. \] Therefore, probability that \(B\) misses is \[ 1-\frac{4}{5}=\frac{1}{5}. \]
Step 2: Understand how \(A\) can win.
Since \(A\) shoots first, \(A\) can win in the following ways:
A hits in the first shot, or both \(A\) and \(B\) miss once and then \(A\) hits, or both miss twice and then \(A\) hits, and so on.
Thus, \[ P(A\text{ wins}) = \frac{3}{5} + \left(\frac{2}{5}\cdot\frac{1}{5}\right)\frac{3}{5} + \left(\frac{2}{5}\cdot\frac{1}{5}\right)^2\frac{3}{5} +\cdots \]
Step 3: Identify the geometric series.
The above expression is a geometric progression with first term \[ \frac{3}{5} \] and common ratio \[ \frac{2}{5}\cdot\frac{1}{5}=\frac{2}{25}. \] Therefore, \[ P(A\text{ wins}) = \frac{3}{5} \left[ 1+\frac{2}{25}+\left(\frac{2}{25}\right)^2+\cdots \right]. \]
Step 4: Sum the infinite geometric series.
Using \[ 1+r+r^2+\cdots=\frac{1}{1-r}, \quad |r|<1, \] we get \[ P(A\text{ wins}) = \frac{3}{5}\cdot \frac{1}{1-\frac{2}{25}}. \] \[ = \frac{3}{5}\cdot \frac{1}{\frac{23}{25}}. \] \[ = \frac{3}{5}\cdot \frac{25}{23}. \] \[ = \frac{15}{23}. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{15}{23}} \]
Probability that \(A\) hits the target is \[ P(A)=0.6=\frac{3}{5}. \] Therefore, probability that \(A\) misses is \[ 1-\frac{3}{5}=\frac{2}{5}. \] Probability that \(B\) hits the target is \[ P(B)=0.8=\frac{4}{5}. \] Therefore, probability that \(B\) misses is \[ 1-\frac{4}{5}=\frac{1}{5}. \]
Step 2: Understand how \(A\) can win.
Since \(A\) shoots first, \(A\) can win in the following ways:
A hits in the first shot, or both \(A\) and \(B\) miss once and then \(A\) hits, or both miss twice and then \(A\) hits, and so on.
Thus, \[ P(A\text{ wins}) = \frac{3}{5} + \left(\frac{2}{5}\cdot\frac{1}{5}\right)\frac{3}{5} + \left(\frac{2}{5}\cdot\frac{1}{5}\right)^2\frac{3}{5} +\cdots \]
Step 3: Identify the geometric series.
The above expression is a geometric progression with first term \[ \frac{3}{5} \] and common ratio \[ \frac{2}{5}\cdot\frac{1}{5}=\frac{2}{25}. \] Therefore, \[ P(A\text{ wins}) = \frac{3}{5} \left[ 1+\frac{2}{25}+\left(\frac{2}{25}\right)^2+\cdots \right]. \]
Step 4: Sum the infinite geometric series.
Using \[ 1+r+r^2+\cdots=\frac{1}{1-r}, \quad |r|<1, \] we get \[ P(A\text{ wins}) = \frac{3}{5}\cdot \frac{1}{1-\frac{2}{25}}. \] \[ = \frac{3}{5}\cdot \frac{1}{\frac{23}{25}}. \] \[ = \frac{3}{5}\cdot \frac{25}{23}. \] \[ = \frac{15}{23}. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{15}{23}} \]
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