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Alchen [17]
3 years ago
11

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.
Mathematics
1 answer:
butalik [34]3 years ago
4 0

Answer:

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

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Alchen [17]

Answer:

The figure PEST is a rhombus

Step-by-step explanation:

* Lets talk about the difference between all these shapes

- At first to prove the shape is a parallelogram you must have one

 of these conditions

# Each two opposite sides are parallel OR

# Each two opposite sides are equal in length OR

# Its two diagonals bisect each other

- After that to prove the parallelogram is:

* A rectangle you must have one of these conditions

# Two adjacent sides are perpendicular to each other OR

# Its two diagonals are equal in length

* A rhombus you must have one of these conditions

# Two adjacent sides are equal in length OR

# Its two diagonals perpendicular to each other OR

# Its diagonals bisect its vertices angles

* A square you must have two of these conditions

# Its diagonals are equal and perpendicular OR

# Two adjacent sides are equal and perpendicular

* Now lets solve the problem

∵ The vertices of the quadrilateral PEST are

   P (-1 , -5) , E (8 , 2) , S (11 , 13) , T (2 , 6)

- Lets find the slope from each two points using this rule :

 m = (y2 - y1)/(x2 - x1), where m is the slope and (x1 , y1) , (x2 , y2)

 are two points on the line

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ m of PE = (2 - -5)/(8 - -1) = 7/9

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ m of ES = (13 - 2)/(11 - 8) = 11/3  

- Let (x1 , y1) is (11 , 13) and (x2 , y2) is (2 , 6)

∴ m of ST = (6 - 13)/(2 - 11) = -7/-9 = 7/9

- Let (x1 , y1) is (2 , 6) and (x2 , y2) is (-1 , -5)

∴ m of TP = (-5 - 6)/(-1 - 2) = -11/-3 = 11/3

∵ m PE = m ST = 7/9

∴ PE // ST ⇒ opposite sides

∵ m ES = m TP = 11/3

∴ ES // TP ⇒ opposite sides

- Each two opposite sides are parallel

∴ PEST is a parallelogram

- Lets check if the parallelogram can be rectangle or rhombus or

 square by one of the condition above

∵ If two line perpendicular , then the product of their slops = -1

- Lets check the slopes of two adjacent sides (PE an ES)

∵ m PE = 7/9

∵ m ES = 11/3

∵ m PE × m ES = 7/9 × 11/3 = 77/27 ≠ -1

∴ PE and ES are not perpendicular

∴ PEST not a rectangle or a square (the sides of the rectangle and

  the square are perpendicular to each other)

- Now lets check the length of two adjacent side by using the rule

 of distance between two points (x1 , y1) and (x2 , y2)

 d = √[(x2 - x1)² + (y2 - y1)²]

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ PE = √[(8 - -1)² + (2 - -5)²] = √[9² + 7²] = √[81 + 49] = √130 units

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ ES = √[(11 - 8)² + (13 - 2)²] = √[3² + 11²] = √[9 + 121] = √130 units

∴ PE = ES ⇒ two adjacent sides in parallelogram

∴ The four sides are equal

* The figure PEST is a rhombus

4 0
3 years ago
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