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

Firstly, the probability that a voter supports the Alliance party is a constant.
Secondly, each voter’s decision to vote for Alliance Party is independent of another voter’s decision.
Lastly, voters can only decide to vote or not vote for Alliance Party.

\mathrm{P}(A = 3) \mathrm{P}(A = 4) = 0.373

Since n is large, np = 16.5 > 5 ~\&~ nq = 13.5 > %, we can approximate A with a normal distribution.

Since n is large, np = 16.5 > 5, we cannot approximate A with a poisson distribution.

\mathrm{P}(A = 15) = 0.06864

{30 \choose 15} p^{15} (1-p)^{15} - 0.06864

[p(1-p)]^{15} = 4.42503 \times 10^{-10}

p(1-p) = 0.2379

p = 0.39 \text{~or~} 0.61

Since p \textless 0.5, \Rightarrow p = 0.39

KS Comments:

(i), writing two of the three will suffice since it is two marks.
(iv), it is necessary that we show that equation so students need to pay attention to their expansion of indices here. The formula applied here can be found in MF15 so students should take note.

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