<p>These are from PR 2009-2010.</p>
<p>Can the following question (#1) can be done by binompdf on the calculator?</p>
<ol>
<li>In a basketball shooting contest, if the probability that Heather will make a basket on any given attempt is 4/5, then what is the probability that she will make at least one basket in three attempts?</li>
</ol>
<p>a. 12/125
b. 64/125
c. 124/125
d. 1
e. 12/5</p>
<p>(Answer is C)</p>
<p>How do you get the answer for the following question WITHOUT plugging in an x value?</p>
<ol>
<li>[x!(x + 1)!]/(x - 1)! =
a. x!(x + 2)!
b. (x^2 - 1)!
c. x(x - 2)!
d. x!(x - 1)
e. x(x + 1)!</li>
</ol>
<p>(Answer is E)</p>
<p>Please explain!
Thanks.</p>
<p>1: use complementary probability. P(make at least one in 3 attempts)+P(0 makes)=1. So, 1-P(0 makes) gives us the desired. The only way to make 0 in three attempts is to miss all 3: 1/5<em>1/5</em>1/5=1/125. so 1-1/125=124/125</p>
<p>2: Simplify. x!(x+1)!/(x-1)!=x!(x-1)!(x)(x+1)/(x-1)!=x!(x)(x+1)=(x+1)!x</p>
<p>1) binomcdf, since it says she will make at least one so you know its not discrete and you won’t use binompdf. binomcdf (3,4/5,1,3)=.992=124/125
Sorry I’m not sure how to do number 2, could someone else please explain.</p>
<p>You cant plug in 4 numbers for binomial cdf. Just do what morpheus says.1- Probability of her not making any.</p>
<p>yes you can binomcdf (n,p, low,high)</p>
<p>@morpheus44, thanks. How did you get to the 2nd step of question 2 though, ie. x!(x-1)!(x)(x+1)/(x-1)! ? How come you can add (x)(x+1) in at the numerator. Sorry if this is a really stupid question haha gah its horrifying but I totally can’t remember math, took my last exam in it last November.</p>
<p>@angelhere, thanks, I tried typing that into my TI-84 and got an error message. Is that the right order/formula, (n, p, min, max)?</p>
<p>i have a ti89 and it works just fine when i type it in, maybe its different for a ti84.</p>
<p>Split (x+1)! into (x+1)<em>(x)</em>(x-1)!</p>