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bixtya [17]
3 years ago
12

Suppose customer arrivals at a post office are modeled by a Poisson process N with intensity λ > 0. Let T1 be the time of the

first arrival. Let t > 0. Suppose we learn that by time t there has been precisely one arrival, in other words, Nt = 1. What is the distribution of T1 under this new information? In other words, find the conditional probability P(T1 ≤ s|Nt = 1) for all s ≥ 0.
Mathematics
1 answer:
Lynna [10]3 years ago
4 0

Answer:

Step-by-step explanation:

We need to find the conditional probability P( T1 < s|N(t)=1 )  for all s ≥ 0

P( time of the first person's arrival < s till time t exactly 1 person has arrived )

= P( time of the first person's arrival < s, till time t exactly 1 person has arrived ) / P(exactly 1 person has arrived till time t )

{ As till time t, we know that exactly 1 person has arrived, thus relevant values of s : 0 < s < t }

P( time of the first person arrival < s, till time t exactly 1 person has arrived ) / P(exactly 1 person has arrived till time t )

= P( exactly 1 person has arrived till time s )/ P(exactly 1 person has arrived till time t )

P(exactly x person has arrived till time t ) ~ Poisson(kt) where k = lambda

Therefore,

P(exactly 1 person has arrived till time s )/ P(exactly 1 person has arrived till time t )

= [ kse-ks/1! ] / [ kte-kt/1! ]

= (s/t)e-k(s-t)

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1 year ago
Can u solve this for me thank u
ser-zykov [4K]

Answer:

Part (A) → x = 5

Part (B) → m(arc RS) = 55°

Step-by-step explanation:

In the picture attached,

RU and SV are the diameters of the given circle.

Part (A),

Since RU is the diameter,

m(arc RVU) = 180°

m(arc RV) + m(arc VU) = 180°

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Part (B),

Since m(arc RS) = m(arc VU) [Angles intercepted by these arcs at the center are equal because they are the vertical angles]

m(arc RS) = (10x + 5)

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4 0
3 years ago
Three different non-zero digits can be arranged in six different ways to
alex41 [277]

Answer:

1134

Step-by-step explanation:

We have 3 digits:

a, b, c

a 3 digit number can be written as:

a*100 + b*10 + c*1

Such that these numbers can be:

{1, 2, 3, 4, 5, 6, 7, 8, 9}

Let's assume that:

a < b < c

Then the 3 smaller numbers are:

a*100 + b*10 + c

a*100 + c*10 + b

b*100 + a*10 + c

The 3 larger numbers are:

b*100 + c*10 + a

c*100 + a*10 + b

c*100 + b*10 + a

We know that the sum of the 3 smaller numbers is equal to 540, then:

(a*100 + b*10 + c) + (a*100 + c*10 + b) + (b*100 + a*10 + c) = 540

Let's simplify this:

(a + a + b)*100 + (b + c + a)*10 + (c + b + c) = 540

(2a + b)*100 + (b + c + a)*10 + (2c + b) = 540

The sum of the 3 larger numbers is equal to X, we want to find the value of X:

(b*100 + c*10 + a) + (c*100 + a*10 + b) + (c*100 + b*10 + a) = X

Now let's simplify the left side:

(b + c + c)*100 + (c + a + b)*10 + (a + b + a)*1 = X

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

Then we have two equations:

(2a + b)*100 + (b + c + a)*10 + (2c + b) = 540

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

Notice that the terms are inverted.

By looking at the first equation, we can see that:

(2c + b) = 10    (because the units digit of 540 is 0)

Then, we can see that:

(b + c + a + 1 ) = 14   (the one comes from the previous 10)

finally:

(2a + b + 1) = 5   (the one comes from the previous 14)

Then we can rewrite:

(2*c + b) = 10

(b + c + a) = 14 -1  = 13

(2a + b) = 5 - 1 = 4

Now we can replace these 3 in the equation:

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

(10)*100 + (13)*10 + 4 = X

1000 + 130 + 4 = X

1134 = X

The sum of the 3 largest numbers is 1134.

6 0
3 years ago
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