All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

Firstly, the probability of picking a free gift is constant. Secondly, there is only two possible outcomes, i.e., the packets either contain a free gift or do no contain a free gift.

F~\sim~B(20, ~0.05 )
\mathrm{P}(F=1) = 0.377

F~\sim~B(60, ~0.05 )
Since n is large, np = 3 < 5, ~ F~\sim~\mathrm{Po}(3) approximately. \mathrm{P}(F >ge 1) = 1 - \mathrm{P}(F \le 4) = 0.185

KS Comments:

This is a simple question on binomial distribution and approximation. Students had to be cautious to answer (i) in context.

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