heureka

A small regional carrier accepted 21 reservations for a particular flight with 18 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the raining passengers will arrive for the flight with a 45% chance, independently of each other.

1.  Find the probability that overbooking occurs

2.  Find the probability that the flight has empty seats

My attempt:

1. Find the probability that overbooking occurs

overbooking mean that more than 7 of the other 10 will arrive for the flight:

probability of 8 arriving is \(10C_8*0.45^8*0.55^2 = 45*0.45^8*0.55^2= 0.02288958944\)
probability of 9 arriving is \(10C_9*0.45^9*0.55^1 = 10*0.45^9*0.55^1= 0.00416174353\)
probability of 10 arriving is \(10C_{10}*0.45^{10}*0.55^0 = 1*0.45^{10}*1= 0.00034050629\)

That sum is 0.02739183926.

The probability that overbooking occurs is \(\mathbf{0.0274}\)

2.  Find the probability that the flight has empty seats

Just 7, all seats are full. The probability is \(10C_7*0.45^7*0.55^3 = 120*0.45^7*0.55^3= 0.07460310632\)
this plus the 0.02739183926 is 0.10199494558.
That is the probability no empty seats will occur.

The probalbility hat empty seats will occur is the compliment \((1-0.10199494558) = 0.89800505442\)

The probability that the flight has empty seats is \(\mathbf{0.8980}\)

Four positive integers \(p,q,r,s\) satisfy the following equations:
\(\begin{align*} pq+2p+q&=348 \\ qr+4q+3r&=373 \\ rs+8r+6s&=544 \end{align*}\)

The system of equations

\(\dfrac{xy}{x+y}=1,\ \dfrac{xz}{x+z}=2,\ \dfrac{yz}{y+z}=3 \)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{xy}{x+y}} &=& \mathbf{1} \\\\ \dfrac{x+y}{xy} &=& 1 \\\\ \dfrac{x}{xy}+\dfrac{y}{xy} &=& 1 \\\\ \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} &=& \mathbf{1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\dfrac{xz}{x+z}} &=& \mathbf{2} \\\\ \dfrac{x+z} {xz}&=& \dfrac{1}{2} \\\\ \dfrac{x} {xz}+\dfrac{z} {xz}&=& \dfrac{1}{2} \\\\ \mathbf{\dfrac{1} {z}+\dfrac{1}{x}} &=& \mathbf{\dfrac{1}{2}} \qquad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\dfrac{yz}{y+z}} &=& \mathbf{3} \\\\ \dfrac{y+z}{yz} &=& \dfrac{1}{3} \\\\ \dfrac{y}{yz}+\dfrac{z}{yz} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} &=& \mathbf{\dfrac{1}{3}} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{x=\ ?} \\ \hline (1)+(2)-(3): & \dfrac{1}{y}+\dfrac{1}{x} + \dfrac{1} {z}+\dfrac{1}{x}- \left( \dfrac{1}{z}+\dfrac{1}{y} \right) &=& 1 +\dfrac{1}{2}- \dfrac{1}{3} \\\\ & \dfrac{2}{x} &=& 1 +\dfrac{1}{2}- \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{x} &=& 6 + 3- 2 \\\\ & \dfrac{12}{x} &=& 7 \\\\ & \mathbf{x} &=& \mathbf{\dfrac{12}{7}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{y=\ ?} \\ \hline (1)-(2)+(3): & \dfrac{1}{y}+\dfrac{1}{x} - \left(\dfrac{1} {z}+\dfrac{1}{x}\right)+ \dfrac{1}{z}+\dfrac{1}{y} &=& 1 -\dfrac{1}{2}+ \dfrac{1}{3} \\\\ & \dfrac{2}{y} &=& 1 -\dfrac{1}{2}+ \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{y} &=& 6 - 3+ 2 \\\\ & \dfrac{12}{y} &=& 5 \\\\ & \mathbf{y} &=& \mathbf{\dfrac{12}{5}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{z=\ ?} \\ \hline -(1)+(2)+(3): & - \left( \dfrac{1}{y}+\dfrac{1}{x} \right) + \dfrac{1} {z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y} &=& -1 +\dfrac{1}{2}+ \dfrac{1}{3} \\\\ & \dfrac{2}{z} &=& -1 +\dfrac{1}{2}+ \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{z} &=& -6 + 3+ 2 \\\\ & \dfrac{12}{z} &=& -1 \\\\ & \mathbf{z} &=& \mathbf{-12} \\ \hline \end{array}\)

Find the coefficient of \(x^3*y*z^2\) in the expandsion of \((3x - 5y + z)^6\).

Multinomial theorem:

For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n:
\(\left( x_1+x_2+\cdots +x_m \right)^n = \sum \limits_{k_1+k_2+\dots+k_m=n} \dbinom{n}{k_1,k_2,\dots,k_m}\cdot x_1^{k_1}\cdot x_2^{k_2}\dots x_m^{k_m}\)
where  
\(\dbinom{n}{k_1,k_2,\dots,k_m} = \dfrac{n!}{k_1!k_2!\dots k_m!}\)

Given that point O is the center of the circle and OA=5, OC=8, and \(\angle ACO=30^\circ\), find \(\angle BOA\) in degrees.

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{\sin\left(90^\circ+\dfrac{x}{2}\right) }{8}} &=& \mathbf{\dfrac{\sin(30^\circ)}{5}} \quad | \quad \sin\left(90^\circ+\dfrac{x}{2}\right) = \cos\left( \dfrac{x}{2} \right),\ \sin(30^\circ) = \dfrac{1}{2} \\\\ \dfrac{\cos\left( \dfrac{x}{2} \right) }{8} &=& \dfrac{\dfrac{1}{2}}{5} \\\\ \dfrac{\cos\left( \dfrac{x}{2} \right) }{8} &=& \dfrac{1}{10} \\\\ \cos\left( \dfrac{x}{2} \right) &=& \dfrac{8}{10} \\\\ \cos\left( \dfrac{x}{2} \right) &=& \dfrac{4}{5} \\\\ \dfrac{x}{2} &=& \arccos\left( \dfrac{4}{5} \right) \\\\ \dfrac{x}{2} &=& 36.8698976458^\circ \\ x &=& 2* 36.8698976458^\circ \\ \mathbf{x} &=& \mathbf{73.7397952917^\circ} \\ \hline \end{array}\)

\(\mathbf{\angle BOA \approx 73.74^\circ}\)

The following grid shows a magic square.  
What is the sum of the three numbers in any row?
\(\begin{array}{c|c|c} 2x & 3 & 2 \\ \hline & & -3 \\ \hline 0 & x & \end{array}\)

Find x:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} 2x & 3 & 2 \\ \hline & & -3 \\ \hline 0 & x & \color{red}y \end{array} \\ 0+x+y &=& 2-3+y \\ x &=& 2-3 \\ \mathbf{x}&=& \mathbf{-1} \\ \hline \end{array}\)

The sum of the three numbers in any row:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} \color{red}-2 & \color{red}3 & \color{red}2 \\ \hline & & -3 \\ \hline 0 & -1 & \end{array} \\ -2+3+2 &=& \mathbf{3} \\ \hline \end{array}\)

Find y:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} -2 & 3 & 2 \\ \hline & \color{red}y& -3 \\ \hline 0 & -1 & \end{array} \\ 0+y+2 &=& 3 \\ y &=& 3-2 \\ \mathbf{y}&=& \mathbf{1} \\ \hline \end{array}\)

Find y:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} -2 & 3 & 2 \\ \hline & 1& -3 \\ \hline 0 & -1 & \color{red}y \end{array} \\ -2+1+y &=& 3 \\ -1+y &=& 3\\ y &=& 3+1\\ \mathbf{y}&=& \mathbf{4} \\ \hline \end{array}\)

Find y:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} -2 & 3 & 2 \\ \hline\color{red}y & 1& -3 \\ \hline 0 & -1 & 4 \end{array} \\ -2+y+0 &=& 3 \\ y &=& 3+2 \\ \mathbf{y}&=& \mathbf{5} \\ \hline \end{array}\)

The magic square:

\(\begin{array}{|rcll|} \hline \begin{array}{c|c|c} -2 & 3 & 2 \\ \hline 5 & 1& -3 \\ \hline 0 & -1 & 4 \end{array} \\ \hline \end{array} \begin{array}{rcll} 0+1+2 &=& 3 \\ -2+3+2 &=& 3 \\ 5+1-3 &=& 3 \\ 0-1+4 &=& 3 \\ -2+1+4 &=& 3 \\ -2+5+0 &=& 3 \\ 3+1-1 &=& 3 \\ 2-3+4 &=& 3 \\ \end{array}\)

We construct the following sequence:
Beginning at 0, we move to the next whole number by changing our first number by one,

our second number by two, our third number by three, etc.
If we are able to move backwards to a whole number that has not been previously listed, we do so, otherwise we move forward.
The first five numbers of this sequence are 0, 1, 3, 6, 2.
Find the 20th number in this sequence.
What is the first number to appear twice in this sequence?

Recaman's sequence:

\(\begin{array}{|r|r|} \hline 1 & 0, \\ & & +1 \\ 2 & 1, \\ & & +2 \\ 3& 3, \\ & & +3 \\ 4& 6, \\ & & -4 \\ 5& 2, \\ & & +5 \\ 6& 7, \\ & & +6 \\ 7& 13, \\ & & +7 \\ 8& 20, \\ & & -8 \\ 9& 12, \\ & & +9 \\ 10& 21, \\ & & -10 \\ 11& 11, \\ & & +11 \\ 12& 22, \\ & & -12 \\ 13& 10, \\ & & +13 \\ 14& 23, \\ & & -14 \\ 15& 9, \\ & & +15 \\ 16& 24, \\ & & -16 \\ 17& 8, \\ & & +17 \\ 18& 25, \\ & & +18 \\ 19& 43, \\ & & +19 \\ \color{red}20& \color{red}62, \\ & & -20 \\ 21& \color{green} 42, \\ & & +21 \\ 22& 63, \\ & & -22 \\ 23& 41, \\ & & -23 \\ 24& 18, \\ & & +24 \\ 25& \color{green} 42, \\ & & -25 \\ 26& 17, \\ & &+26 \\ 27& 43, \\ \ldots & \ldots \\ \hline \end{array}\)

In a University out of 120 students, 15 opted mathematics only, 16 opted statistics only, 9 opted physics only and
45 opted physics and mathematics, 30 opted physics and statistics, 8 opted mathematics and statistics, and
80 opted physics.
Find the sum of number of students who opted mathematics and those who didn't opted any of the subjects given.

My attempt:

\(\begin{array}{|rcll|} \hline x+y+t+9 &=& 80 \quad | \quad x+y = 45 \\ 45+t+9 &=& 80 \\ t+54 &=& 80 \\ t &=& 80-54 \\ \mathbf{ t } &=& \mathbf{26} \\ \hline \end{array} \begin{array}{|rcll|} \hline 30 &=& y+t \\ -~~8 &=& y+z \\ \hline 22 &=& y+t-(y+z) \\ 22 &=& y+t-y-z \\ 22 &=& t-z \\ z+22 &=& t \\ \mathbf{ z } &=& \mathbf{t-22} \quad | \quad \mathbf{ t =26} \\ z &=& 26-22 \\ \mathbf{ z } &=& \mathbf{4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline z+y &=& 8 \\ y &=& 8-z \quad | \quad \mathbf{z=4} \\ y &=& 8-4\\ \mathbf{ y } &=& \mathbf{4} \\ \hline \end{array} \begin{array}{|rcll|} \hline x+y &=& 45 \\ x &=& 45-y \quad | \quad \mathbf{y=4} \\ x &=& 45-4\\ \mathbf{ x } &=& \mathbf{41} \\ \hline \end{array}\)

The sum of number of students who opted mathematics:

\(\begin{array}{|rcll|} \hline \text{Mathematics} &=& 15+x+y+z \\ \text{Mathematics} &=& 15+41+4+4 \\ \mathbf{\text{Mathematics}} &=& \mathbf{64} \\ \hline \end{array} \)

The sum of number of students who didn't opted any of the subjects given:

\(\begin{array}{|rcll|} \hline \text{didn't opted any} &=& 120-(9+15+16+x+y+z+t) \\ \text{didn't opted any} &=& 120-(40+41+4+4+26) \\ \text{didn't opted any} &=& 120-115 \\ \mathbf{\text{didn't opted any}} &=& \mathbf{5} \\ \hline \end{array}\)

During a football game, Matt kicked the ball three times.
His longest kick was 43 metres and the three kicks averaged 37 metres.
If the other two kicks were the same length, the distance, in metres, that each travelled was what?

\(\begin{array}{|rcll|} \hline \dfrac{43+2x}{3} &=& 37 \\\\ 43+2x &=& 37*3 \\ 43+2x &=& 111 \\ 2x &=& 111-43 \\ 2x &=& 68 \\\\ 2x &=& \dfrac{68}{2} \\\\ \mathbf{ x } &=& \mathbf{34}\ \text{metres} \\ \hline \end{array}\)

The lengths of the sides of a triangle are 5, 6 and 10 cm respectively.

Find the square of the length of the median of the greatest side.

cos-rule:

\(\begin{array}{|lrcll|} \hline (1) & 6^2 &=& 5^2+10^2-2*5*10*\cos(A) \\ & 36 &=& 125-100\cos(A) \\ & 100\cos(A) &=& 125-36 \\ & 100\cos(A) &=& 89 \\ &\mathbf{ \cos(A) } &=& \mathbf{\dfrac{89}{100}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (2) & x^2 &=& 5^2+5^2-2*5*5*\cos(A) \\ & x^2 &=& 50-50\cos(A) \\ & x^2 &=& 50-50*\dfrac{89}{100} \\ & x^2 &=& 50-\dfrac{89}{2} \\ & x^2 &=& 50-44.5 \\ &\mathbf{ x^2 } &=& \mathbf{5.5} \\ \hline \end{array}\)

The square of the length of the median of the greatest side is \(\mathbf{5.5}\ \text{cm}\)