People are arriving at a party one at a time. while waiting for more people to arrive they entertain themselves by comparing their birthdays. let x be the number of people needed to obtain a birthday match, i. e., before person x arrives there are no two people with the same birthday, but when person x arrives there is a match. find the pmf of x.

=(k−1)*P(X>k−1) or (k−1)365k(365k−1)(k−1)!

Step-by-step explanation:

First of all, we need to find PMF

Let X = k represent the case in which there is no birthday match within (k-1) people

However, there is a birthday match when kth person arrives

Hence, there is 365^k possibilities in birthday arrangements

Supposing (k-1) dates are placed on specific days in a year

Pick one of k-1 of them & make it the date of the kth person that arrives, then:

The CDF is P(X≤k)=(1−(365k)k)/!365k, so the can obtain the PMF by  

P(X=k) =P (X≤k) − P(X≤k−1)=(1−(365k)k!/365^k)−(1−(365k−1)(k−1)!/365^(k−1))=

(k−1)/365^k * (365k−1) * (k−1)!

=(k−1)*(1−P(X≤k−1))

=(k−1)*P(X>k−1)

b

step-by-step explanation:

b

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}