Consider the population of voters described in example 3.6. suppose that there are n = 5000 voters in the population, 40% of whom favor jones. identify the event favors jones as a success s. it is evident that the probability of s on trial 1 is .40. consider the event b that s occurs on the second trial. then b can occur two ways: the first two trials are both successes or the first trial is a failure and the second is a success. show that p(b) = .4. what is p(b| the first trial is s)? does this conditional probability differ markedly from p(b)?

Step-by-step explanation:

We know that P( S ) = 0.4

Probability of occurance of event B can be calculated as follows

This probability consists of two elements

1 ) Probability of first two trial becoming successful

= .4 x .4 = .16

2 )  a )    Probability  of first trial becoming failure =  1-.4

= 0.6

b ) probability of second trial becoming success = .4

Probability of occurance of both a ) and b ) event simultaneously one after another.

= 0.6 x 0.4

.24

Total probability of .first two trials becoming  both successes or the first trial is a failure and the second is a success is

.16+ .24 = .4

Hence

P(B) = .4

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

⭐⭐

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}