Question

A spinner is divided into 6 equal parts. These sections are labeled with a number 1, one section has a 2, one section has a 3, and one section has a 4. If the spinner is spun two times, what is the probability it will land on both 2 sides?

Answer 1

Answer:

$\frac{1}{36} = 0.0278$ = 2.78% probability it will land on both 2 sides.

Step-by-step explanation:

For each trial, there are only two possible outcomes. Either it lands on 2, or it does not. The probability of the spinner landing on two in a trial is independent of any other trial, which means that the binomial probability distribution is used 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.

One of the six sides is 2:

This means that $p = \frac{1}{6}$

If the spinner is spun two times, what is the probability it will land on both 2 sides?

This is P(X = 2) when n = 2. So

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

$P(X = 2) = C_{2,2}.(\frac{1}{6})^{2}.(\frac{5}{6})^{0} = \frac{1}{36} = 0.0278$

$\frac{1}{36} = 0.0278$ = 2.78% probability it will land on both 2 sides.