# Spinning a wheel, the probability of landing in area A is .99 probability of landing in area B is .01. In 60 spins what is the probability of landing in area B?

Jun 4, 2017

Probability = 0.6%

#### Explanation:

First we convert the probability of Area B from dec
To find the probability of landing in Area B, we multiply the probability of area B by the number of spins, so it looks like
$P = 0.01 \cdot 60$, where P is the probability of landing in Area B. Do the math, and the answer becomes $P = 0.6$. This doesn't look like the answer until you realize that it means $P = \frac{0.6}{100}$. Knowing that $\frac{6}{100}$ converted to percentage = 6%, and that $\frac{0.6}{100}$ is 6% divided by 10, one can see how (6%)/10=0.6%, which makes 0.6/100=0.6%.

I hope that helped!

Jun 4, 2017

$P \left(\text{at least 1 B}\right) = 0.4528$

#### Explanation:

There are only two possible sets of outcomes:

$\text{all 60 spins land on A}$

$\text{at least one spin lands on B}$

Therefore, we can say that:

$P \left(\text{all 60 A") + P("at least 1 B}\right) = 1$

So in order to find the chance of at least 1 spin landing on B, let's first find the chance that all 60 spins land on A:

$P \left(\text{all 60 A") = underbrace(P(A)timesP(A)times cdots timesP(A))_(60 color(white)"." "times}\right) = {\left(P \left(A\right)\right)}^{60}$

$= {0.99}^{60} \approx 0.5472$

Therefore:

$P \left(\text{all 60 A") + P("at least 1 B}\right) = 1$

$0.5472 + P \left(\text{at least 1 B}\right) = 1$

$P \left(\text{at least 1 B}\right) = 0.4528$