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3 years ago

Some cows are brown. Fido is not a cow. Fido must be brown.

Mathematics

2 answers:

Lina20 [59]3 years ago

8 0

Answer:

B.) Invalid Argument

Step-by-step explanation:

Just because Fido is not a cow does not mean that he will automatically be brown. It is already stated that some cows(something Fido is not) are colored brown.

astraxan [27]3 years ago

5 0

Answer:

B invalid

Step-by-step explanation:

just because fido is not a cow doesn't mean he is brown...fido could be any other colour

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A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic

goldfiish [28.3K]

Answer:

(a) The proportion of women who are tested, get a negative test result is 0.82.

(b) The proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event E $_{i}$ , of the sample space S, given that another event A has already occurred is:

$P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}$

The law of total probability states that, if events E₁, E₂, E₃... are parts of a sample space then for any event A,

$P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}$

Denote the events as follows:

X = fetus have a chromosome abnormality.

Y = the test is positive

The information provided is:

$P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85$

Using the above the probabilities compute the remaining values as follows:

$P(X^{c})=1-P(X)=1-0.04=0.96$

$P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10$

$P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.85=0.15$

(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

$P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})$

$=(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82$

Thus, the proportion of women who are tested, get a negative test result is 0.82.

(b)

Compute the value of P (X|Y) as follows:

Use the Bayes' theorem:

$P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}$

$=\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}$

$=0.20$

Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

6 0

3 years ago

Simplify (4ab^2)^3 Any help is appreciated

Pie

Answer:

$64a^{3} b^{6}$

Step-by-step explanation:

1) Distribute the exponent outside of the parentheses to the terms inside the parentheses. In this case, multiply the 3 outside the parentheses with the exponent of each of the terms inside:

$(4ab^{2} )^3$

$4^{3}$ · $a^{3}$ · $b^{6}$

2) The variables are fine - you can't simplify them further. All you need to do now is multiply the $4^{3}$ out. Remember that it's kind of like asking, "what is 4 × 4 ×4?"

$4$ · $4$ · $4$ · $a^{3}$ · $b^{6}$