The probability that three men hit a target are respectively 1.docx
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The document calculates the probability that the first of three men hits a target when only one of them successfully hits it. It details the probabilities of each man hitting and missing the target and derives the final result, which shows the probability that the first man hit the target given that only one hit it. The final probability is calculated to be 6/31.
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The probability that three men hit a target are respectively 1.docx
- 1. The probability that three men hit a target are respectively 1/6, ¼ and 1/3. Each shoots once at the target. If only one hit the target, What is the probability that it was the first man? Solution 𝑴 = 𝑯𝒊𝒕 𝒕𝒉𝒆 𝒕𝒂𝒓𝒈𝒆𝒕 𝑴 ̅ = 𝑵𝒐𝒕 𝒉𝒊𝒕 𝒕𝒉𝒆 𝒕𝒂𝒓𝒈𝒆𝒕 𝑷(𝑴𝟏) = 𝟏 𝟔 𝑷(𝑴 ̅ 𝟏) = 𝟓 𝟔 𝑷(𝑴𝟐) = 𝟏 𝟒 𝑷(𝑴 ̅ 𝟐) = 𝟑 𝟒 𝑷(𝑴𝟑) = 𝟏 𝟑 𝑷(𝑴 ̅ 𝟑) = 𝟐 𝟑 𝑷(𝑶𝒏𝒍𝒚 𝒐𝒏𝒆 𝒎𝒂𝒏 𝒉𝒊𝒕 𝒕𝒉𝒆 𝒕𝒂𝒓𝒈𝒆𝒕) = 𝑷(𝑯) 𝑷(𝑯) = 𝑷(𝑴𝟏) × 𝑷(𝑴 ̅ 𝟐) × 𝑷(𝑴 ̅ 𝟑) + 𝑷(𝑴 ̅ 𝟏) × 𝑷(𝑴𝟐) × 𝑷(𝑴 ̅ 𝟑) + 𝑷(𝑴 ̅ 𝟏) × 𝑷(𝑴 ̅ 𝟐) × 𝑷(𝑴𝟑) 𝑷(𝑯) = 𝟏 𝟔 × 𝟑 𝟒 × 𝟐 𝟑 + 𝟓 𝟔 × 𝟏 𝟒 × 𝟐 𝟑 + 𝟓 𝟔 × 𝟑 𝟒 × 𝟏 𝟑
- 2. 𝑷(𝑯) = 𝟔 + 𝟏𝟎 + 𝟏𝟓 𝟕𝟐 = 𝟑𝟏 𝟕𝟐 𝑷(𝑭𝒊𝒓𝒔𝒕 𝒎𝒂𝒏 𝒉𝒊𝒕 𝒕𝒉𝒆 𝑯 ⁄ ) = 𝑷(𝑭 𝑯 ⁄ ) 𝑷(𝑭 𝑯 ⁄ ) = 𝑷(𝑴𝟏) × 𝑷(𝑴 ̅ 𝟐) × 𝑷(𝑴 ̅ 𝟑) 𝑷(𝑯) 𝑷(𝑭 𝑯 ⁄ ) = 𝟏 𝟔 × 𝟑 𝟒 × 𝟐 𝟑 𝟑𝟏 𝟕𝟐 = 𝟔 𝟑𝟏