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?

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*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= 0.6/100`. Knowing that `6/100` converted to percentage = 6%, and that `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!

There are only two possible sets of outcomes:

`"all 60 spins land on A"`

`"at least one spin lands on B"`

Therefore, we can say that:

`P("all 60 A") + P("at least 1 B") = 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("all 60 A") = underbrace(P(A)timesP(A)times cdots timesP(A))_(60 color(white)"." "times") = (P(A))^60`

`=0.99^60 ~~ 0.5472`

Therefore:

`P("all 60 A") + P("at least 1 B") = 1`

`0.5472 + P("at least 1 B") = 1`

`P("at least 1 B") = 0.4528`

Final Answer