Question 1147534
<i>(a) Calculate the probability that no more than 4 of them will eat pizza for lunch.</i>
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Probability exactly four will eat pizza: {{{(0.80)^4 * (0.20)^4 * (8!/(4!*4!))}}} = 0.0459
Probability exactly three will eat pizza: {{{(0.80)^3 * (0.20)^5 * (8!/(3!*5!))}}} = 0.0092
Probability exactly two will eat pizza: {{{(0.80)^2 * (0.20)^6 * (8!/(2!*6!))}}} = 0.0011
Probability exactly one will eat pizza: {{{(0.80)^1 * (0.20)^7 * (8!/(1!*7!))}}} = 0.0001
Probability none will eat pizza: {{{(0.20)^8}}} = 0.000003
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Probability no more than four will eat pizza for lunch: {{{0.0459 + 0.0092 + 0.0011 + 0.0001 + 0.000003}}} = 0.0563 (rounded)
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<i>(b) Calculate the probability that exactly 2 of them will eat chicken nuggets for lunch.</i>
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{{{(0.20)^2 * (0.80)^6 * (8!/(2!*6!))}}} = 0.2936
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<i>(c) Calculate the probability that none of them will eat pizza for lunch.</i>
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We already calculated this in Part A.
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{{{(0.20)^8}}} = 0.000003