Question

A spinner has 12 equal sectors. 4 sectors are coloured red, 3 are coloured blue, and 5 are coloured yellow. The pointer on the spinner is spun 3 times. What is the probability of the pointer landing on red each time?

Answer 1

Answer:

3.70% probability of the pointer landing on red each time

Step-by-step explanation:

For each time that the pointer is spun, there are only two possible outcomes. Either it lands on red, or it does not. The probability that it lands on red on a spin is independent of other spins. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

12 sections, of which 4 are red.

This means that $p = \frac{4}{12} = 0.3333$

The pointer on the spinner is spun 3 times.

This means that $n = 3$

What is the probability of the pointer landing on red each time?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.3333)^{3}.(0.6667)^{0} = 0.0370$

3.70% probability of the pointer landing on red each time