Question

Geologist has collected 10 specimens of basaltic rock and 10th specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. What is the probability of getting exactly 5 granite specimens selected for analysis(a) What is the pmf of the number of granite specimens selected for analysis?
(b) What is the probability that all specimens of one of the two types of rock are selected
for analysis?
(c) What is the probability that the number of granite specimens selected for analysis is
within 1 standard deviation of its mean value?

Answer 1

Answer:

$p(5)=0.00136$

(a) $p(x)=\frac{(10Cx)*((10)C(15-x))}{20C10}$

(b)$p(10)+p(5)=0.00272$

(c)$p(7)+p(8)=0.0584$

Step-by-step explanation:

If we have N elements with k elements that we consider success and N-k elements that we consider fail, and we take a sample of n elements, the probability that there are x successes in the sample follows a hypergeometric distribution, so it is calculated as:

$p(x)=\frac{(kCx)*((N-k)C(n-x))}{NCn}$

In this case, N is equal to 20, k is equal to 10 specimens of granite rock, N-k is equal to 10 specimens of basaltic rock and n is 15. So, the probability that there are x specimens of granite rock in the sample or the pmf of the number of granite specimens selected for analysis is:

$p(x)=\frac{(10Cx)*((10)C(15-x))}{20C10}$

Now, the probability of getting exactly 5 granite specimens selected for analysis is:

$p(5)=\frac{(10C5)*((10)C(15-5))}{20C10}=0.00136$

On the other hand, the probability that all specimens of one of the two types of rock are selected  for analysis is the probability to select 10 granite specimens or 10 basaltic specimens. Where the probability to take 10 basaltic specimens is equal to the probability to take 5 granite specimens.

Then, the probability that all specimens of one of the two types of rock are selected  for analysis is:

$p(10)+p(5)=\frac{(10C10)*((10)C(15-10))}{20C10}+\frac{(10C5)*((10)C(15-5))}{20C10}\\p(10)+p(5)=0.00136+0.00136=0.00272$

For the hypergeometric distribution, the mean E(x) and standard deviation S(x) are:

$E(x)=\frac{nk}{N}=\frac{15*10}{20}=7.5\\S(x)=\sqrt{\frac{nk}{N}*(1-\frac{k}{N})*(\frac{N-n}{N-1})}\\S(x)=\sqrt{\frac{15*10}{20}*(1-\frac{10}{20})*(\frac{20-15}{20-1})}=0.9934$

Then, the number of granite specimens within 1 standard deviation of its mean value is:

  E(x) - S(x) ≤ mean ≤ E(x) + S(x)

7.5-0.9934 ≤ mean ≤ 7.5+0.9934

     6.5066 ≤ mean ≤ 8.4934

               7 ≤ mean ≤ 8

Finally, the probability that the number of granite specimens selected for analysis is  within 1 standard deviation of its mean value is:

$p(7)+p(8)=\frac{(10C7)*((10)C(15-7))}{20C10}+\frac{(10C8)*((10)C(15-8))}{20C10}\\p(10)+p(5)=0.0292+0.0292=0.0584$