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sleet_krkn [62]
2 years ago
13

Which conditional probabilities are correct? check all that apply. p(d | f) = startfraction 6 over 34 endfraction p(e | d) = sta

rtfraction 7 over 25 endfraction p(d | e) = startfraction 7 over 25 endfraction p(f | e) = startfraction 8 over 18 endfraction p(e | f) = startfraction 13 over 21 endfraction
Mathematics
1 answer:
Bas_tet [7]2 years ago
6 0

The conditional probabilities that are correct are P(D | F) =  6/34, P(E | D) = 7/25 and P(F | E) = 8/18

<h3>How to determine the true conditional probabilities ?</h3>

The formula to compute the conditional probability P(A|B) is:

P(A | B) = P(A and B)/P(B)

The above means that the probability of event A such that the event B has already occurred

When the above formula is applied to the give data in the complete, we have:

P(D | F) = 6/34

P(E | D) = 7/25

P(D | E) = 7/18

P(F | E) = 8/18

P(E | F) = 8/34

Hence, the conditional probabilities that are correct are P(D | F) =  6/34, P(E | D) = 7/25 and P(F | E) = 8/18

See attachment for complete question

Read more about conditional probabilities at:

brainly.com/question/10739997

#SPJ4

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Let X equal the number of typos on a printed page with a mean of 4 typos per page.
timama [110]

Answer:

a) There is a 98.17% probability that a randomly selected page has at least one typo on it.

b) There is a 9.16% probability that a randomly selected page has at most one typo on it.

Step-by-step explanation:

Since we only have the mean, we can solve this problem by a Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

In this problem, we have that \mu = 4

(a) What is the probability that a randomly selected page has at least one typo on it?

Thats is P(X \geq 1). Either a number is greater or equal than 1, or it is lesser. The sum of the probabilities must be decimal 1. So:

P(X < 1) + P(X \geq 1) = 1

P(X \geq 1) = 1 - P(X < 1)

In which

P(X < 1) = P(X = 0).

So

P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183

P(X \geq 1) = 1 - P(X < 1) = 1 - 0.0183 = 0.9817

There is a 98.17% probability that a randomly selected page has at least one typo on it.

(b) What is the probability that a randomly selected page has at most one typo on it?

This is P = P(X = 0) + P(X = 1). So:

P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183

P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733

P = P(X = 0) + P(X = 1) = 0.0183 + 0.0733 = 0.0916

There is a 9.16% probability that a randomly selected page has at most one typo on it.

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