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lesya692 [45]
3 years ago
13

If I chose a number uniformly from the integers from 1 to 25, calculate the conditional probability that the number is a multipl

e of 6 (including 6) given that it is larger than 18.
Mathematics
1 answer:
topjm [15]3 years ago
7 0

Answer:

<h2>1/7</h2>

Step-by-step explanation:

If I choose a number from the integers 1 to 25, the total number of integers I can pick is the total outcome which is 25. n(U) = 25

Let the probability that the number chosen at random is a multiple of  6 be P(A) and the probability that the number chosen at random is is larger than 18 be P(B)

P(A) = P(multiple of 6)

P(B) = P(number larger than 18)

A = {6, 12, 18, 24}

B = {19, 20, 21, 22, 23, 24, 25}

The conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18 is expressed as P(A|B).

P(A|B) = P(A∩B)/P(B)

Since probability = expected outcome/total outcome

A∩B = {24}

n(A∩B) = 1

P(A∩B) = n(A∩B)/n(U)

P(A∩B) = 1/25

Given B = {19, 20, 21, 22, 23, 24, 25}.

n(B) = 7

p(B) = n(B)/n(U)

p(B) = 7/25

Since P(A|B) = P(A∩B)/P(B)

P(A|B) = (1/25)/(7/24)

P(A|B) = 1/25*25/7

P(A|B) = 1/7

<em></em>

<em>Hence the conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18 is 1/7</em>

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The maximum value of f(x) occurs at:

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And is given by:

\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

Step-by-step explanation:

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Step-by-step explanation:

We are given the function:

\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

\displaystyle f(0) = (0)^a(2-(0))^b = 0

And:

\displaystyle f(2) = (2)^a(2-(2))^b = 0

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]

By the Product Rule:

\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}

Set the derivative equal to zero and solve for <em>x: </em>

\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]

By the Zero Product Property:

\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0

The solutions to the first equation are <em>x</em> = 0 and <em>x</em> = 2.

First, for the second equation, note that it is undefined when <em>x</em> = 0 and <em>x</em> = 2.

To solve for <em>x</em>, we can multiply both sides by the denominators.

\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))

Simplify:

\displaystyle a(2-x) - b(x) = 0

And solve for <em>x: </em>

\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\  \frac{2a}{a+b} &= x  \end{aligned}

So, our critical points are:

\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

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This can be simplified to:

\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

Since <em>a</em> and <em>b</em> > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let <em>a</em> = 2 and <em>b</em> = 3. Then:

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The critical point will be at:

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Testing <em>x</em> = 0.5 and <em>x</em> = 1 yields that:

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Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

\displaystyle x = \frac{2a}{a+b}

And is given by:

\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

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