Which of the following are equivalent to 5 4/6 ?

For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

In this problem

Of the 20 copies, 2 are defective, so $p = \frac{2}{20} = 0.1$ .

What is the probability that you will encounter neither of the defective copies among the 10 you examine?

This is P(X = 0) when $n = 10$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.3487$

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

8 0

3 years ago

Of all the soft drink consumers in a particular sales region, 30% prefer Brand A and 70% prefer Brand

victus00 [196]

Answer:

Option D -0.67

Step-by-step explanation:

Given : Preference of Brand A =30% ⇒ $P(A)=\frac{30}{100}=0.3$

Preference of Brand B =70% ⇒ $P(B)=\frac{70}{100}=0.7$

Prefer Brand A and are Female = 20% ⇒ $P(A/F)=\frac{20}{100}=0.2$

Prefer Brand B and are Female = 40% ⇒ $P(B/F)=\frac{40}{100}=0.4$

To find : Selected consumer is female, given that the person prefers Brand A $P(F/A)$

Solution : Using Bayes' theorem, which state that

$P(A/B)=\frac{P(B/A)P(A)}{P(B)}$

where, P(A) and P(B) are probabilities of observing A and B.

P(B/A)= is a conditional probability where event B occur and A is true

P(A/B)= also a conditional probability where event A occur and B is true.

Now, applying Bayes' theorem,

$P(F/A)=\frac{P(A/F)P(A)}{P(B)P(B/F)+P(A)P(A/F)}$

$P(F/A)=\frac{(0.2)(0.3)}{(0.7)(0.4)+(0.3)(0.2)}$

$P(F/A)=\frac{0.6}{0.28+0.6}$

$P(F/A)=\frac{0.6}{0.88}$

$P(F/A)=0.68$

Therefore, Option D is correct probability that a randomly selected consumer is female, given that the person prefers Brand A -0.67

3 0

3 years ago

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