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A spinner at a school fair has eight equal sectors, and exactly one of them is gold…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A spinner at a school fair has eight equal sectors, and exactly one of them is gold. A visitor spins it four times. Find the probability that gold never shows, and the probability that gold shows at least once. Then find the smallest number of spins that makes the probability of at least one gold greater than $0.5$ .

Need a reminder? The key ideaShow

For equally likely outcomes, $P(E) = \dfrac{n(E)}{n(S)}$ . In general the probabilities of the simple events are at least $0$ and add up to $1$ , and $P(E') = 1 - P(E)$ . For at least one, use $1 - P(\text{none})$ . Odds of $a : b$ in favour mean $P = \dfrac{a}{a + b}$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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