1. Find the conditional probability of the indicated event when two fair dice (one red and one green) are rolled. The sum is 6, given that the green one is either 4 or 1. 2. Find the conditional probability of the indicated event when two fair dice (one red and one green) are rolled. The red one is 6, given that the sum is 11.

Answer 1:

Conditional probability Formula :

P (A/B) = P (A∩B)/ P (B).

Total sample space=36 outcomes

Conditions are :

The probability of (A∩B)= P (A∩B)= 2/36= 1/18

Now we find the probability of green die having 4 or 1

When green is 4 red can have all the numbers from 1- 6

And when green is 1 red can have all the numbers from 1- 6

Total number = 12

 P (B)= 12/36

Therefore, conditional probability = P (A/B)

The conditional probability of the indicated event when two fair dice  are rolled will be 1/6.

Answer 2:

Condition :

When red is 6 the green must be 5 to get 11.

P (A∩B)=  1/36

The probability of red die having 6 =P(B)= 6/36

The conditional probability= P (A∩B)/P(B)

The conditional probability of the indicated event when two fair dice are 1/6.

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1. 1/6

2. 1/6

Step-by-step explanation:

Let A be the event that the sum of the two die is 6 and B be an event that the green die is either 4 or 1.

The conditional probability will be given by P (A/B) = P (A∩B)/ P (B).

Now the total sample space consists of 36 outcomes .

And to find (A∩B) we need to find the outcomes in which green die is either 4 or 1 and the sum of the two die is 6.

So when green is 1 red must be 5

So when green is 4 red must be 2

So there are two ways in which green die is either 4 or 1 and the sum of the two die is 6.

Therefore the probability of (A∩B)= P (A∩B)= 2/36= 1/18

Now we find the probability of green die having 4 or 1

So when green is 4 red can have all the numbers from 1- 6

And when green is 1 red can have all the numbers from 1- 6

The total number would be 12 .

So probability of green die having 1 or 4 is given by = P (B)= 12/36

Now the conditional probability = P (A/B) = P (A∩B)/ P (B)=1/18/ 1/3

= 3/18= 1/6

2. Similarly we find the conditional probability of the two die when the red one is 6, given that the sum is 11.

When red is 6 the green must be 5 to get 11. So the probability

=P (A∩B)=  1/36

Now we find the probability of red die having 6 =P(B)= 6/36

Now the conditional probability = P (A∩B)/P(B) =  1/36/ 6/36= 1/6

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}

can you rephrase the question ? it is unclear : -)

step-by-step explanation: