a. $0.216$ b. $0.064$ c. $0.936$
Work Step by Step
a. All have assistantships $p=0.6^3=0.216$ b. None has an assistantship $p=(1-0.6)^3=0.064$ c. At least 1 has an assistantship This is the complement of case (b) $p=1-0.064=0.936$
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