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3 years ago

Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

Mathematics

1 answer:

Marina86 [1]3 years ago

7 0

Answer:

The probability of selecting a black card or a 6 = 7/13

Step-by-step explanation:

In this question we have given two events. When two events can not occur at the same time,it is known as mutually exclusive event.

According to the question we need to find out the probability of black card or 6. So we can write it as:

P(black card or 6):

The probability of selecting a black card = 26/52

The probability of selecting a 6 = 4/52

And the probability of selecting both = 2/52.

So we will apply the formula of compound probability:

P(black card or 6)=P(black card)+P(6)-P(black card and 6)

Now substitute the values:

P(black card or 6)= 26/52+4/52-2/52

P(black card or 6)=26+4-2/52

P(black card or 6)=30-2/52

P(black card or 6)=28/52

P(black card or 6)=7/13.

Hence the probability of selecting a black card or a 6 = 7/13 ....

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3 years ago

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

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3 years ago

(A) How many ways are there to distribute seven identical apples and six identical pears to three distinct people such that each

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Answer:

Step-by-step explanation:

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4 years ago

While looking at some reports, a store manager notes that televisions that retail for $475 and up costing customers $513 once th

Mekhanik [1.2K]

The sales tax percentage of the store manager report is 7.41%

<h3>How to solve for the sales tax percentage</h3>

Customers are subject to a charge known as sales tax when they buy goods and services.

It is a pass-through tax, which means you must collect it from clients and send the money to your state or local government. the seller do not contribute sales tax.

The sales tax is first calculated by

= price after tax - price before tax

= 513 - 475

= 38

sales tax percentage is calculated using the formula

= (Tax amount / Price before tax) × 100%

= (38 / 513) * 100

= 7.4074

= 7.41%

Learn more about sales tax at:

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