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victus00 [196]
3 years ago
6

f two eventsAandBare independent, then we knowP(A∩B) =P(A)P(B). A fact is that ifAandBare independent, then so are all combinati

ons ofA,B,...etc.Show that if eventsAandBare independent, thenP(A∩B) =P(A)P(B), and thusAandBare independent. (Hint:P(A∩B) = 1−P(A∪B). Then use addition rule and simplify.)
Mathematics
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

B^c and A^c are independent

Step-by-step explanation:

The proof:

We assume that A and B are independent.

By definition, the occurrence of A doesn't affect the probability of B.

Thus, the occurrence of A also doesn't affect the probability of B^c

So by definition, A and B^c are also independent, which by definition again means that the occurrence of B^c doesn't affect the probability of A.

(Here we used the symmetry of independence.)

Therefore, the occurrence of B^c also doesn't affect the probability of A^c.

So by definition, B^cand A^c are also independent.

One could convert the proof to the language of math:

A and B are independent↓P(B|A)=P(B)↓1−P(B^c|A)=1−P(B^c)↓P(B^c|A)=P(B^c)↓A and B^c are independent↓P(A|B^c)=P(A)↓1−P(A^c|B^c)=1−P(A^c)↓P(A^c|B^c)=P(A^c)↓B^c and A^c are independent But now we used conditional probabilities, which might be a problem in case P(A)=0 or P(B^c)=0

OR

Since,

P(A') = 1-P(A)

P(B') = 1- P(B)

Now clearly P(A′)P(B′)=1−[P(A)+P(B)]+P(A∩B)

From set algebra we know that P(A)+P(B)=P(A∪B)+P(A∩B) Substituting, we have P(A′)P(B′)=P([A∪B]′)

Now from De morgans law we know that:

[A∪B]′=[A′∩B′]

Substituting, we have P(A′)P(B′)=P(A′∩B′) , as required.

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7, 10, 13

Step-by-step explanation:

1) n = 1:   3(1) + 4 = 3 + 4 = 7

2) n = 2:   3(2) + 4 = 6 + 4 = 10

3) n = 3:   3(3) + 4 = 9 + 4 = 13

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The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 104 inches, and a standard
dsp73

Answer:

91.92% probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 104, \sigma = 14, n = 49, s = \frac{14}{\sqrt{49}} = 2

What is the probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

This is the pvalue of Z when X = 106.8. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{106.8 - 104}{2}

Z = 1.4

Z = 1.4 has a pvalue of 0.9192

0.9192 = 91.92% probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

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