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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: pinklemonade on August 14, 2014, 07:43:03 pm

Title: Quick Methods Question
Post by: pinklemonade on August 14, 2014, 07:43:03 pm: I just need help with this multiple choice question on bionomial distribution. I know the answer is A but I was wondering if anyone could just show the working out for it please!

If X is a random variable, binomially distributed with n=10 and p=k, $Pr(X\geq 1)$ is:

$a.) 1-(1-k)^{10}$

$b.) (1-k)^{10}$

$c.) 10(k)(1-k)^{9}$

$d.) (k)^{10}$

$e.) 1-(k)^{10}$
Title: Re: Quick Methods Question
Post by: paper-back on August 14, 2014, 08:39:01 pm: $Pr(x\geq 1)=_{1}^{10}\textrm{C}(k)^{1}(1-k)^{9}+_{2}^{10}\textrm{C}(k)^{2}(1-k)^{8}+... \\\text{As this is tedious,(unless using BinomialCDF) a better way to do it would be like this; } \\\text{Remember: All probabilites add up to 1 } \\\therefore Pr(x\geq 1)=_{1}^{10}\textrm{C}(k)^{1}(1-k)^{9}+_{2}^{10}\textrm{C}(k)^{2}(1-k)^{8}+...=\text{(is the same as)} \\1-_{0}^{10}\textrm{C}(k)^{0}(1-k)^{10} \\\therefore 1-(1-k)^{10}$
Title: Re: Quick Methods Question
Post by: pinklemonade on August 14, 2014, 09:08:23 pm: Quote from: paper-back on August 14, 2014, 08:39:01 pm
$Pr(x\geq 1)=_{1}^{10}\textrm{C}(k)^{1}(1-k)^{9}+_{2}^{10}\textrm{C}(k)^{2}(1-k)^{8}+... \\\text{As this is tedious,(unless using BinomialCDF) a better way to do it would be like this; } \\\text{Remember: All probabilites add up to 1 } \\\therefore Pr(x\geq 1)=_{1}^{10}\textrm{C}(k)^{1}(1-k)^{9}+_{2}^{10}\textrm{C}(k)^{2}(1-k)^{8}+...=\text{(is the same as)} \\1-_{0}^{10}\textrm{C}(k)^{0}(1-k)^{10} \\\therefore 1-(1-k)^{10}$

Thank you! It makes sense now :)
Title: Re: Quick Methods Question
Post by: pinklemonade on August 19, 2014, 06:59:23 pm: I have another quick probability question which I'm having trouble with!

An unbiased 8-sided die is rolled 12 times. The probability of obtaining three results greater than 5 is:
a.) 0.1135
b.) 0.1688
c.) 0.2188
d.) 0.2279
e.) 0.2824
Title: Re: Quick Methods Question
Post by: keltingmeith on August 19, 2014, 07:25:08 pm: Quote from: pinklemonade on August 19, 2014, 06:59:23 pm
I have another quick probability question which I'm having trouble with!

An unbiased 8-sided die is rolled 12 times. The probability of obtaining three results greater than 5 is:
a.) 0.1135
b.) 0.1688
c.) 0.2188
d.) 0.2279
e.) 0.2824

The fact that there are 8 outcomes would probably make you think that isn't binomial - but, it is. In this case, we'll define a success as any result greater than 5. This means that we can now define the random variable X such that $X\sim Bi\left(12,\frac{3}{8}\right)$

Now, we want to know the probability that exactly 3 results are greater than 5 - that is, 3 successes. So, by our handy-dandy formula, we get $^{12}C_3 \left(\frac{3}{8}\right)^3\left(\frac{5}{8}\right)^9$ , which (using an online calculator because I don't know where my CAS is) is 0.1688, b.
Title: Re: Quick Methods Question
Post by: pinklemonade on August 19, 2014, 07:46:49 pm: Quote from: EulerFan101 on August 19, 2014, 07:25:08 pm
The fact that there are 8 outcomes would probably make you think that isn't binomial - but, it is. In this case, we'll define a success as any result greater than 5. This means that we can now define the random variable X such that $X\sim Bi\left(12,\frac{3}{8}\right)$

Now, we want to know the probability that exactly 3 results are greater than 5 - that is, 3 successes. So, by our handy-dandy formula, we get $^{12}C_3 \left(\frac{3}{8}\right)^3\left(\frac{5}{8}\right)^9$ , which (using an online calculator because I don't know where my CAS is) is 0.1688, b.

Thank you! Helped alot