Do you have an equal chance of landing in either 1 or 4?

Mathematics

1 answer:

vitfil [10]3 years ago

4 0

Answer:

Yes

Step-by-step explanation:

Since the circle is split into 6 pieces, we can split the probability of landing on each section into fractions

There is one piece with the number 1 on it, so there is a 1/6 chance of landing on that.

There are 3 pieces with the number 2 on it, so there is a 3/6 chance of landing on it, simplifies that would be 1/2

There is one piece with the number 3 on it, so there is a 1/6 chance of landing on that.

There is one piece with the number 4 on it, so there is a 1/6 chance of landing on that.

Since the pieces with a 1 and 4 on the both have a 1/6 probabilit of landing, there is an equal chance on landing on either tile.

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Express each of the following problems algebraically

serg [7]

Sent a picture of the solutionsolution to the problem (s).

7 0

4 years ago

There are five seniors in a class. For each situation, write how the binomial formula is used to calculate the probability.

ololo11 [35]

A) The answer is 5.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
<span>I choose one senior: r = 1
</span>
nCr = n! / (r! (n - r)!)
5C1 = 5! / (1! (5 - 1)!)
= (5 * 4 * 3 * 2 * 1) / (1 * 4!)
= 120 / (4 * 3 * 2 * 1)
= 120 / 24
= 5

b) The answer is 10.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose two seniors: r = 2

nCr = n! / (r! (n - r)!)
5C2 = 5! / (2! (5 - 2)!)
= (5 * 4 * 3 * 2 * 1) / ((2 * 1) * 3!)
= 120 / (2 * (3 * 2 * 1))
= 120 / (2 * 6)
= 120 / 12
= 10

c) The answer is 10.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose three seniors: r = 3

nCr = n! / (r! (n - r)!)
5C3 = 5! / (3! (5 - 3)!)
= (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * 2!)
= 120 / (6 * (2 * 1))
= 120 / (6 * 2)
= 120 / 12
= 10

d) The answer is 5.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose four seniors: r = 4

nCr = n! / (r! (n - r)!)
5C4 = 5! / (4! (5 - 4)!)
= (5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * 1!)
= 120 / (24 * 1)
= 120 / 24
= 5

e) The answer is 1.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose five seniors: r = 5

nCr = n! / (r! (n - r)!)
5C5 = 5! / (5! (5 - 5)!)
= (5 * 4 * 3 * 2 * 1) / ((5 * 4 * 3 * 2 * 1) * 1!)
= 120 / (120 * 1)
= 120 / 120
= 1

8 0

4 years ago

Simplify 5(y + 6) + 15y

KATRIN_1 [288]

Answer:

Step-by-step explanation: