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Musya8 [376]
3 years ago
15

Need a quick answer please

Mathematics
1 answer:
mars1129 [50]3 years ago
8 0

Answer:

Step-by-step explanation:

Y = 1 because if you have 1/x that whould meant it would be 1 so y equals 1

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A manager forecasts that 150 guests will be served tomorrow. The manager knows that one server can provide high quality service
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6 0
3 years ago
jon flips a fair coin until a head is witnessed, and independently anna flips a fair coin (a different coin than jon’s) until a
Readme [11.4K]

a) The pmf for w for certain value n is P(W = n) = 3 / 2²ⁿ⁺², b) The expected value on w, E(w) = 4 / 3, c) The expression for P(w ≥ k) =  \frac{1}{2}^{2k}, where k = 1, 2, 3...

Given: Jon flips a fair coin until a head is witnessed.

Anna flips a fair coin independently until a head is witnessed.

What is the probability density function?
The probability density function is a function that determines the probability for a discrete random variable X over a given sample space S for a certain value for X = x(some value).

It is also denoted as pmf and is written as P(X = x).

Let's solve the given question:

Let us assume the number of flips Jon did until he got a head to be X.

Also as Anna's flip independently so we need to consider a different random variable say Y.

So, the probability mass function for Jon (X) for a certain value of x is:

P(X = x) = \frac{1}{2} * (1 - \frac{1}{2})^{x-1}, where x = 1, 2, 3, ...........

P(X = x) = \frac{1}{2}*\frac{1}{2}^{x - 1}

P(X = x) = \frac{1}{2}^{x - 1 + 1}

P(X = x) = \frac{1}{2}^x , where x = 1, 2, 3, ........

Similarly, for Anna, we will have the same probability mass function but with a different random variable Y for a certain value of y

P(Y = y) = \frac{1}{2}^y, where y = 1, 2, 3, ........

Now it is given that, w is the minimum number of flips between Jon and Anna.

w = minimum(P(X = x), P(Y = y))

Let us suppose the probability mass distribution over w is n for a certain value.

Then

P(w ≥ n) = P(minimum(P(X = x), P(Y = y)) ≥ n)

= \frac{1}{2}^n * \frac{1}{2}^n

= \frac{1}{2}^{2n}

Therefore, the probability mass function for w is

P(W = n) = P(W ≥ n) - P(W ≥ n + 1)

= \frac{1}{2}^{2n}\\ - \frac{1}{2}^{2(n+1)}

= \frac{1}{2}^{2n} - \frac{1}{2}^{2n + 2}

= \frac{1}{2}^{2n}( 1 - \frac{1}{2}^{2} )

= \frac{1}{2}^{2n} ( 1 - 1 / 4)

= \frac{1}{2}^{2n}(4 - 1) / 4

= 3 / 4 (\frac{1}{2}^{2n})

= 3 / 2² (\frac{1}{2}^{2n})

= 3 / 2²ⁿ⁺²

Now the expected value of w, E(w) is:

E(w) = ∑P(w ≥ i) where i = 0 to ∞

= ∑\frac{1}{2^{2i}}<em> </em>where i = 0 to ∞

= 1 + 1 / 4 + 1 / 16 + .......

This is infinite GP series. So the summation of infinite GP is

S = a / ( 1 - r )  

where a is the first term, r is the power and s is the summation.

Here a = 1, r = 1 / 4

S = 1 / (1 - 1 / 4)

S = 1 / 3 / 4

S = 4 / 3

Therefore, E(w) = 4 / 3

The expression for P(w ≥ k) is:

P(w ≥ k) =  \frac{1}{2}^{2k}, where k = 1, 2, 3...

Hence

a) The pmf for w for certain value n is P(W = n) = 3 / 2²ⁿ⁺²

b) The expected value on w, E(w) = 4 / 3

c) The expression for P(w ≥ k) =  \frac{1}{2}^{2k}, where k = 1, 2, 3...

Know more about "probability density function" here: brainly.com/question/14410995

#SPJ4

5 0
2 years ago
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