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

(i)
(a)
Let X be the number of cakes containing fruit, out of 6.

X ~\sim~ \mathrm{B} (6, 0.4)

\mathrm{P} (X = 0) = 0.0467

(b)
\mathrm{P} (X \le 2) = 0.544

(ii)
Let Y be the number of packets with at most 2 cakes containing fruit, out of 8.

Y ~\sim~ \mathrm{B} (8, 0.54432)

\mathrm{P} (Y \ge 4) = 1 - \mathrm{P}(Y \le 3) = 0.729

(iii)
Let W be the number of packets with at most 2 cakes containing fruit, out of 150.

W ~\sim~ \mathrm{B}(150, 0.54432)

Since n = 150 is large, np = 81.648 > 5 and nq = 68.352 > 5,

W ~\sim~ \mathrm{N}(81.648, 37.20536) approximately.

\mathrm{P}(W > 75) = \mathrm{P}(W > 75.5) by continuity correction
= 0.843

KS Comments

Students must call all variables out carefully as they denote different things. And many still forget to perform continuity correction.

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