Binomial distributions

Question

(x+y)^5
(4x+y)^4
(2x+y)^5
(n+2m)^4
At hopewell high school, 1 in 7 students is on a sports team. There are 4 student council representatives in the school. What is the probability that 2 of the student council representatives are on a sports team? What is the probability that at least 3 of te student council representatives are on a sports team? Two in every 5 donuts in a package of donuts have a jelly filling. There are 5 donuts left in the package. What is the probability that all 5 donuts have a jelly filling? What is the probability that none of the donuts has a jelly filling? Andrew is choosing CDs from a bag of free CDs without looking. He has a 1 in 5 chance of choosing a cd that he likes. He chooses 8 CDs in all. What is the probability that he will get 3 CDs that he likes? In a game of bingo, the contestants have a 1 in 12 chance of winning each round. If Shirley plays 6 rounds, what is the probability that she will win at least half of them?

Answer 1

There are a lot of questions here, so I am going to start with the first four that deal with binomials. A binomial raised to a power simply means that you would have to write out the binomial four times and multiply it through. Luckily, we can use the BINOMIAL THEOREM. This theorem uses the rows of Pascals Triangle to determine the coefficients. For the first one:

 

(x+y)⁵= x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵

 

I will jump to the third example since it will take the same form as the first, we will just substitute in with the different values, in this case x=2x and y=y:

 

(2x+y)⁵ = 2x⁵ + 10x⁴y + 20x³y² + 20x²y³ + 10xy⁴ + y⁵

 

The other two will take the form of a binomial raised to the fourth power, which takes the general form of:

 

(x+y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

 

Given the second problem, the form would become:

 

(4x+y)⁴ = 4x⁴ + 16x³y + 24x²y² + 16xy³ + y⁴

 

The final problem would simplify to:

 

(n+2m)⁴ = n⁴ + 8n³m + 12n²m² + 8nm³ + 2m⁴

 

Hopefully that helps for those first four problems. Remember, the powers of each term in the extended formula should always add up to the power the binomial is being raised to.

 

Answer 2

(2x+y)^5
= (C(5,0)) ((2x)^5) ((y)^0)
+ (C(5,1)) ((2x)^4) ((y)^1)
+ (C(5,2)) ((2x)^3) ((y)^2)
+ (C(5,3)) ((2x)^2) ((y)^3)
+ (C(5,4)) ((2x)^1) ((y)^4)
+ (C(5,5)) ((2x)^0) ((y)^5)

Where C(n,r) = n!/(r!(n-r)!) and is the r'th Pascal's Triangle Element in the n'th row, where both indices start at 0.

Pascal's Triangle
1
1 1
1 2   1
1 3   3   1
1 4   6   4   1
1 5 10 10   5 1
1 6 15 20 15 6 1

(2x+y)^5
= (1) (2^5)(x^5) (1)
+ (5) (2^4)(x^4) (y)
+ (10) (2^3)(x^3) (y^2)
+ (10) (2^2)(x^2) (y^3)
+ (5) (2x) (y^4)
+ (1) (1) (y^5)

(2x+y)^5
= (1) (32)(x^5) (1)
+ (5) (16)(x^4) (y)
+ (10) (8)(x^3) (y^2)
+ (10) (4)(x^2) (y^3)
+ (5) (2x) (y^4)
+ (1) (1) (y^5)

(2x+y)^5
= 32x^5
+ 80x^4 y
+ 80x^3 y^2
+ 40x^2 y^3
+ 10x y^4
+ y^5

(2x+y)^5 =
32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5

===== I think the following are right; I followed the examples on http://www.mathsisfun.com/data/binomial-distribution.html. I'm looking forward to seeing other tutors’ solutions.

“At Hopewell High School, 1 in 7 students is on a sports team. There are 4 student council representatives in the school. What is the probability that 2 of the student council representatives are on a sports team?”

(1/7)(1/7)(6/7)(6/7)(4*3)/(2*1) = (6^2)*2*3/7^4 = 6^3/7^4 ≈ 0.08996251561849 ≈ 9.00 %

“What is the probability that at least 3 of the student council representatives are on a sports team?”

((1/7)^3)(6/7)*4*3*2/(3*2*1) = 24/7^4 ≈ 0.00999583506872 ≈ 1.000 %

=====

“Two in every 5 donuts in a package of donuts have a jelly filling. There are 5 donuts left in the package. What is the probability that all 5 donuts have a jelly filling?”

((2/5)^5)*5*4*3*2*1/(5*4*3*2*1) = (2/5)^5 = 0.01024 = 1.024 %

What is the probability that none of the donuts has a jelly filling?”

100 % - 1.024 % = 98.976 %

=====

“Andrew is choosing CDs from a bag of free CDs without looking. He has a 1 in 5 chance of choosing a CD that he likes. He chooses 8 CDs in all. What is the probability that he will get 3 CDs that he likes?”

((1/5)^3)((4/5)^5)*8*7*6/(3*2*1) = ((1/5)^3)((2^2/5)^5)(2^3)(7)
= 7(2^(13))/(5^8) = 7(2^(5))((2/5)^8) = 7(32)((2/5)^8)
= 224((2/5)^8) = 14.680064 %

=====

“In a game of bingo, the contestants have a 1 in 12 chance of winning each round. If Shirley plays 6 rounds, what is the probability that she will win at least half of them?”

((1/12)^3)((11/12)^3)6*5*4/(3*2*1) = 20(11^3)/(12^6) ≈ 0.891498413923 %