# Thread: compute probability

1. ## compute probability

I need to work this math question, using the binomial probability formula,can someone help me figure it out. A 10-question multiple choice exam is given, and each question has five possible answers. Pascal takes his exam and guesses at every question. Use the binomial orobability formula to find the probability(to 5 deciminal places) that:
a) he gets exactly 2 questions correct.
b)he gets no questions correct.
c) he gets at least 1 question correct(use the information from part (b) to answer this part.
d) he gets at least 9 questions correct.
e) Without using the binomial probability formula, determine the probability that he gets exactly 2 questions correct.
f) Compare your answers to parts (a) and (f). If they are not the same, explain why.
Is there a math wiz out there, that could possibly help me with this problem, or try to walk me through it,I would really appreciate it, Thanks Toni

2. Originally Posted by toni_45miss@hotmail.com
I need to work this math question, using the binomial probability formula,can someone help me figure it out. A 10-question multiple choice exam is given, and each question has five possible answers. Pascal takes his exam and guesses at every question. Use the binomial orobability formula to find the probability(to 5 deciminal places) that:
a) he gets exactly 2 questions correct.
b)he gets no questions correct.
c) he gets at least 1 question correct(use the information from part (b) to answer this part.
d) he gets at least 9 questions correct.
e) Without using the binomial probability formula, determine the probability that he gets exactly 2 questions correct.
f) Compare your answers to parts (a) and (f). If they are not the same, explain why.
Is there a math wiz out there, that could possibly help me with this problem, or try to walk me through it,I would really appreciate it, Thanks Toni
A binomialy distributed random variable with $N$ trials and probability of a favourable outcome on a single trial $p$ has prbability of $n$ favourable outcomes in $N$ trials of:

$
b(n;N,p) = \frac{N!}{(N-n)!n!}~p^n(1-p)^{n-1}
$

where the exclamation mark denotes the factorial function.

In this question $N=10$ and $p=1/5$, so for part (a) we have:

$
p(n=2) = b(2;10,0.2) = \frac{10!}{8!2!}~0.2^2~0.8^{8}=45 ~0.2^2~0.8^{8} \approx 0.30199
$

RonL

3. ## To Captain Black man you are good, you sure your not in my math class? So do I use 0'

Man you are great, are you sure you are not in my math class? So do I use all 0's for (b),or not. This type of math has lost me
Originally Posted by CaptainBlack
A binomialy distributed random variable with $N$ trials and probability of a favourable outcome on a single trial $p$ has prbability of $n$ favourable outcomes in $N$ trials of:

$
b(n;N,p) = \frac{N!}{(N-n)!n!}~p^n(1-p)^{n-1}
$

where the exclamation mark denotes the factorial function.

In this question $N=10$ and $p=1/5$, so for part (a) we have:

$
p(n=2) = b(2;10,0.2) = \frac{10!}{8!2!}~0.2^2~0.8^{8}=45 ~0.2^2~0.8^{8} \approx 0.30199
$

RonL

4. Hello, toni!

I will assume you know the Binomial Probability Formula.

We know that: . $\begin{array}{ccccc}P(\text{right}) & = & \frac{1}{5} & = & 0.2 \\ P(\text{wrong}) & = & \frac{4}{5} & = & 0.8\end{array}$

A 10-question multiple choice exam is given, and each question has 5 possible answers.
Pascal takes his exam and guesses at every question.
Use the binomial orobability formula to find the probability(to 5 deciminal places) that:

a) he gets exactly 2 questions correct.
$P(\text{2 right})\;=\; {10\choose2}(0.2)^2(0.8)^8$

b) he gets no questions correct.
$P(\text{0 right}) \;=\;{10\choose0}(0.2)^0(0.8)^{10}$

c) he gets at least 1 question correct
(use the information from part (b) to answer this part.
The opposite of "at least one right" is "none right".

Hence: . $P(\text{at least 1 right}) \;=\;1 - P(\text{0 right})$

d) he gets at least 9 questions correct.
"At least 9 right" means: 9 right or 10 right.

$P(\text{at least 9 right}) \;=\;P(\text{9 right}) + P(\text{10 right})$

. . . . . . . . . . . . $= \;{10\choose9}(0.2)^9(0.8)^1 + {10\choose10}(0.2)^{10}(0.8)^0$

I'll let you do the arithmetic . . .

5. Originally Posted by toni_45miss@hotmail.com
I need to work this math question, using the binomial probability formula,can someone help me figure it out. A 10-question multiple choice exam is given, and each question has five possible answers. Pascal takes his exam and guesses at every question. Use the binomial orobability formula to find the probability(to 5 deciminal places) that:
a) he gets exactly 2 questions correct.
b)he gets no questions correct.
c) he gets at least 1 question correct(use the information from part (b) to answer this part.
d) he gets at least 9 questions correct.
e) Without using the binomial probability formula, determine the probability that he gets exactly 2 questions correct.
f) Compare your answers to parts (a) and (f). If they are not the same, explain why.
Is there a math wiz out there, that could possibly help me with this problem, or try to walk me through it,I would really appreciate it, Thanks Toni
Originally Posted by CaptainBlack
A binomialy distributed random variable with $N$ trials and probability of a favourable outcome on a single trial $p$ has prbability of $n$ favourable outcomes in $N$ trials of:

$
b(n;N,p) = \frac{N!}{(N-n)!n!}~p^n(1-p)^{n-1}
$

where the exclamation mark denotes the factorial function.

In this question $N=10$ and $p=1/5$, so for part (a) we have:

$
p(n=2) = b(2;10,0.2) = \frac{10!}{8!2!}~0.2^2~0.8^{8}=45 ~0.2^2~0.8^{8} \approx 0.30199
$

RonL
For part b) we want $n=0$, so:

$
p(n=0) = b(0;10,0.2) = \frac{10!}{10!~0!}~0.2^0~0.8^{10}=0.8^{10} \approx 0.10737
$

Note: $0!=1$

RonL

6. ## Math,math,math!!!

Yes, puzzeled I am, but you guys are explaining it to me in a way I am kinda getting it, thank you all for all of the help, Toni Hopefully someday I can return the favor.