Question

Saul is playing a game using a wheel divided
into two equal sectors, A and B, as shown in the
diagram below. If Saul spins this wheel twice,
what is the probability he will land on A on the
first spin and A and the second?

Answer 1

Using the binomial distribution, it is found that there is a 0.25 = 25% probability he will land on A on the first spin and A and the second.

For each spin, there are only two possible outcomes, either it lands on Sector A, or it does not. The results of each spin are independent, hence, the binomial distribution is used to solve this question.

What is the binomial distribution formula?

The formula is:

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem, we have that:

• The sectors are equal, which means that it is equally as likely to land on each, hence $p = \frac{1}{2} = 0.5$.

• Saul will spin the wheel twice, hence $n = 2$.

The probability that both land on A is P(X = 2), hence:

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

$P(X = 2) = C_{2,2}.(0.5)^{2}.(0.5)^{0} = 0.25$

0.25 = 25% probability he will land on A on the first spin and A and the second.

You can learn more about the binomial distribution at https://brainly.com/question/24863377