The diagram shows a lucky draw spinner at a departmental store.
The pointer is equally likely to stop at any of the sectors.
The sectors show a GRAND prize, 5or5 or 10 prize vouchers to be won or a MISS.
Each customer at the store is entitled to 1 spin for every $50 spent.

  1. (a)

    Find the probability that a customer wins the grand prize in a spin.

    Answer ...

  2. (b)

    Find the probability that a customer wins a 5ora5 or a 10 voucher in a spin.

    Answer ...

  3. (c)

    Mrs Singh spends $120 at the store.
    Find, as a fraction in its simplest form, the probability that she wins at least a prize.

    Answer ...

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Worked solution

19(a)

1. Find the probability of landing on a GRAND sector.

There are 22 GRAND sectors out of 88 equal sectors.

P(GRAND)=28=14P(\text{GRAND})=\frac{2}{8}=\frac{1}{4}

Answer

14\dfrac{1}{4}

19(b)

1. Find the probability of landing on a \$5 or \$10 voucher sector.

There are 33 sectors labelled $5 and 22 sectors labelled $10, out of 88 equal sectors.

P($5 or $10 voucher)=3+28=58P(\text{\$5 or \$10 voucher})=\frac{3+2}{8}=\frac{5}{8}

Answer

58\dfrac{5}{8}

19(c)

1. Find the number of spins Mrs Singh gets.

Mrs Singh spends $120\char36{}120.

Since each spin is given for every $50\char36{}50 spent,
120÷50=2 remainder 20120 \div 50 = 2\text{ remainder }20
So Mrs Singh gets 22 spins.

2. Find the probability that she wins at least one prize in 2 spins.

A prize means any sector except MISS.

For one spin,
P(MISS)=18P(\text{MISS})=\frac{1}{8}

So for 2 spins,
P(no prize in both spins)=18×18=164P(\text{no prize in both spins})=\frac{1}{8}\times\frac{1}{8}=\frac{1}{64}

Hence,
P(at least one prize)=1−164=6364P(\text{at least one prize})=1-\frac{1}{64}=\frac{63}{64}

Answer

6364\dfrac{63}{64}

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