A biased five-sided spinner is numbered 1, 2, 3, 4 and 5 - OCR - GCSE Maths - Question 13 - 2021 - Paper 1

Let the probability of landing on 3 be denoted as $p$. The problem states that the spinner is four times more likely to land on 5 than on 3. Therefore, the probability of landing on 5, $P(5)$, can be represented as:

$P(5) = 4p$

The sum of all the probabilities must equal 1:

$P(1) + P(2) + P(3) + P(4) + P(5) = 1$

Substituting the known probabilities, we have:

$0.10 + 0.10 + p + 0.20 + 4p = 1$

Simplifying: $0.40 + 5p = 1$

Subtract 0.40 from both sides: $5p = 0.60$

Now, divide by 5: $p = 0.12$

Therefore, the probability of landing on 3 is 0.12.