Can someone please help

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

Urn A contains five white balls and seven black balls. Urn B contains three white balls and four black balls. A ball is drawn fr

12345 [234]

Answer:

0.636

Step-by-step explanation:

Let Aw, Ab, Bw, Bb the following events:

Aw = the ball drawn from urn A is white

Ab = the ball drawn from urn A is black

Bw = the ball drawn from urn B is white

Bb = the ball drawn from urn B is black

the probability that the transferred ball was black given that the second ball drawn was black is P(Ab | Bb)

By the Bayes' Theorem

$\bf P(Ab | Bb)=\frac{P(Bb|Ab)P(Ab)}{P(Bb|Ab)P(Ab)+P(Bb|Aw)P(Aw)}=\frac{(5/8)(7/12)}{(5/8)(7/12)+(4/8)(5/12)}$

Working out the calculations

$\bf P(Ab | Bb)=\frac{P(Bb|Ab)P(Ab)}{P(Bb|Ab)P(Ab)+P(Bb|Aw)P(Aw)}=\frac{(5/8)(7/12)}{(5/8)(7/12)+(4/8)(5/12)}=\\\frac{35/96}{35/96+20/96}=35/55=7/11=0.636$

rounded to three decimals

4 0

3 years ago