Spencer is carrying out a survey of the bear population at yellowstone national park. he spots 2 bears - one has a light colored coat and the other has a dark coat.

1- assume that there are equal numbers of male and female bears in the park. what is the probability that both bears are male?
2- if the lighter colored bear is a male, what are the odds that both are male?

1. 1/4

2. 1/2

Step-by-step explanation:

If there are equal numbers of males and females, we can simplify that by saying there's one male and one female for each coat color.

1. What's the probability that both are males?

Well, there's a 1/2 chances the light-colored coat bear is a male and there's also a 1/2 chances the dark-colored coat bear is a male.  So, we can make this reference table:

Light             Dark

Male             Male

Male             Female

Female         Male

Female         Female

As you can see, the probability that both of the spotted bears are males is 1/4 (1/2 * 1/2).

2. If the light-colored bear, what are the odds both are males?

If we know for sure the light-one is male, then there's 1/2 chances the other is too, refer to the table above to verify.

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}