Question 1157344
<i><b>a. all will recover</b></i>
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{{{(0.8)^5}}} = <b>0.32768</b>
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<i><b>b. exactly 2 will recover</b></i>
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{{{(0.8)^2 * (0.2)^3 * (5!/(2!*3!))}}} = {{{(0.8)^2 * (0.2)^3 * 10}}} = 0.64 * 0.008 * 10 = <b>0.0512</b>
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<i><b>c. at least 2 will recover</b></i>
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Find the probability that exactly one will recover AND the probability that none will recover.  Add these together and subtract that result from 1.
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<i>Probability exactly one will recover:</i>
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{{{(0.8)^1 * (0.2)^4 * (5!/(1!*4!))}}} = {{{(0.8)^1 * (0.2)^4 * 5}}} = 0.8 * 0.0016 * 5 = 0.0064
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<i>Probability none will recover:</i>
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{{{(0.2)^5}}} = 0.00032
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<i>Probability exactly one will recover PLUS probability none will recover:
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0.0064 + 0.00032 = 0.00672
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Probability at least two will recover:
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1 - 0.00672 = <b>0.99328</b>