<p>1.a paricular firestation says that the probability that an incoming call is for medical assistance is .85
A. Give a relative frequency interpretation.</p>
<p>E.Assuming independence calculate the probability that exactly one of the next two calls will be for medical assistance.</p>
<p>2.30% of calls to an airline reservation phone line result in a reservation being made.
C.what is the probablility that at least one call results in a reservation being made if 10 calls are made.</p>
<p>Some one explain it is for a take home test</p>
<p>1E= .85*.15 I know this is right, but I'm not great at explaining.
2C= prob. of AT LEAST ONE = 1-prob. of NONE so, 1-(.60^10)</p>
<p>i was having problem with E bc D said calculate probability that for 2 successive calls the first is for medical and the second is not. So that would be the same answer for both right? and .60^10 not .70^10?</p>
<p>okay E is not .85<em>.15.
That would be two consecutive calls that the first is med and second is not as you said.
If I am not mistaken you have to use the formula for a binomial distribution which is P(X)=nCX</em>P^X<em>Q^(n-X) where n=2 and X=1.
Then you have 2C1</em>.85<em>.15. Which equals 2</em>.85*.15.
I might be overthinking this because I don't know how far along you are.<br>
And it is .70^10</p>
<p>yeah, oops on E
.60 was a typo :)</p>
<p>well i am only on chapter 6 so i don't know if we learned that yet. but it is better then what i have which is nothing so i will take it. most people bomb the take home test anyway. so it is worth the try. Thanks you guys</p>