Question

A geologist has collected 5 specimens of basaltic rock and 7 specimens of granite. The geologist instructs a laboratory assistant to randomly select 9 of the specimens for analysis.
Let X= the number of granite specimens selected for analysis.
Note: Take 10 decimal places after the ".", if the answer is a fraction, enter the fraction (a/b). Use a period (.) not a comma (,) for decimals.
a) Compute EX = ?; Var(X) = ?
b) Compute P(X<6) = ?
c) What is the probability that all specimens of one of the two types of rock are selected for analysis?
P(all specimens of one of the two types of rock are selected for analysis) = ?

Answer 1

The rocks are chosen without replacement, which means that the hypergeometric distribution is used to solve this question. First we get the parameters, and then we answer the questions. From this, we get that:

• $E(X) = 5.25, Var(X) = 0.5966$
• P(X < 6) = 0.9545
• P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The mean and the variance are:

$\mu = \frac{nk}{N}$

$\sigma^2 = \frac{nk(N-k)(N-n)}{N^2(N-1)}$

We have that:

5 + 7 = 12 rocks, which means that $N = 12$

9 are chosen, which means that $n = 9$

7 are granite, which means that $k = 7$

Question a:

$E(X) = \mu = \frac{9\times7}{12} = 5.25$

$Var(X) = \sigma^2 = \frac{9\times7(12-7)(12-9)}{12^2(12-1)} = 0.5966$

Thus:

$E(X) = 5.25, Var(X) = 0.5966$

Question b:

Since there are only 5 specimens of basaltic rock, at least 9 - 5 = 4 specimens of granite are needed, which means that:

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,12,9,7) = \frac{C_{7,4}*C_{5,5}}{C_{12,9}} = 0.1591$

$P(X = 5) = h(5,12,9,7) = \frac{C_{7,5}*C_{5,4}}{C_{12,9}} = 0.4773$

$P(X = 6) = h(6,12,9,7) = \frac{C_{7,6}*C_{5,3}}{C_{12,9}} = 0.3181$

Thus

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.1591 + 0.4773 + 0.3181 = 0.9545$

So P(X < 6) = 0.9545.

Question c:

5 of basaltic and 4 of granite: 0.1591 probability.

7 of granite is P(X = 7), in which

$P(X = 7) = h(7,12,9,7) = \frac{C_{7,7}*C_{5,2}}{C_{12,9}} = 0.0455$

0.1591 + 0.0455 = 0.2046, thus:

P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

A similar question is found at brainly.com/question/24008577