Question 1157343
<i><b>(i) if each fires twice, what is the probability that the target will be hit at least once?</b></i>
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Find the probability the target will not be hit at all, then subtract this from 1.
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<i>Probability the target will not be hit at all:</i>
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{{{(3/4)^2 * (7/8)^2}}} = {{{(9/16) * (49/64)}}} = {{{441/1024}}}
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<i>The probability the target will be hit at least once:</i>
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{{{1 - 441/1024}}} = <b>583/1024</b>
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<i><b>(ii) if each fire once and the target is hit only once, what is the probability that A hits the target?</b></i>
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<i>Probability A hits the target and B misses:</i>
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{{{(1/4) * (7/8)}}} = {{{7/32}}}
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<i>Probability B hits the target and A misses:</i>
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{{{(1/8) * (3/4)}}} = {{{3/32}}}
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Probability the "hit" is made by A:
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{{{(7/32)/(7/32 + 3/32)}}} = {{{(7/32)/(10/32)}}} = <b>7/10</b>
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<i><b>(iii) If A can fire only twice, how many times must B fire so that there is at least a 90% probability that the target will be hit?</b></i>
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We want to find how many times B must fire to have at most a 10% probability the target will be missed.
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{{{(3/4)^2 * (7/8)^n}}} ≤ {{{1/10}}}
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{{{(9/16) * (7/8)^n}}} ≤ {{{1/10}}}
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{{{(7/8)^n}}} ≤ {{{(16/9)(1/10)}}}
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{{{(7/8)^n}}} ≤ {{{16/90}}}
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{{{(7/8)^n}}} ≤ {{{8/45}}}
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n = log(base 7/8) 8/45
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n = 12.93
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You want to round this up, so <b>B will have to shoot 13 times to ensure the chance of the target being hit is 90%.</b>