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Probability and Quadratic Equations

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Probability And Quadratic Equations – Introduction

Quadratic equations are a fundamental concept in mathematics, often encountered in the GCSE Higher Mathematics Paper. At the same time, probability is a crucial topic that helps us understand and predict uncertain events. In this section, we will explore the relationship between probability and quadratic equations, highlighting their practical applications.

Probability of Solutions:
 When solving quadratic equations, one important question is: what is the probability of finding a solution? In most cases, quadratic equations have two solutions, but there are instances where they may have one or no real solutions. Therefore, the probability of obtaining real solutions depends on the discriminant of the quadratic equation. If the discriminant is positive, there are two real solutions with a probability of 1. If the discriminant is zero, there is one real solution with a probability of 1. If the discriminant is negative, there are no real solutions, and the probability is 0.

Application in Real-life Scenarios:
 Quadratic equations and probability can be applied to real-life scenarios. For example, when predicting the trajectory of a projectile, such as a ball thrown into the air, we can use quadratic equations to calculate its maximum height, time of flight, and range. Additionally, probability can be used to determine the chances of a certain event occurring, such as winning a lottery or passing an exam. By understanding the relationship between probability and quadratic equations, students can apply these concepts to solve various practical problems.

Probability And Quadratic Equations - Examples

There are initially (n+7) and on the second selection there will be (n+6)

 

\begin{aligned} \mathrm{P}(\mathrm{RR})=\frac{4}{15} & \Rightarrow \frac{n}{n+7} \times \frac{n-1}{n+6}=\frac{4}{15} \\ & \Rightarrow 15 n(n-1)=4(n+7)(n+6) \\ & \Rightarrow 15 n^2-15 n=4 n^2+52 n+168 \\ & \Rightarrow 11 n^2-67 n-168=0 \\ & \Rightarrow(11 n+21)(n-8)=0 \\ & \Rightarrow n=-\frac{21}{11} \text { or } 8, \end{aligned}

So there must be 8 discs in the bag to begin with. 

You need to consider all possibilities. P(RB) and P(BR)

 

\mathrm{P}(\text { one of each colour })=2 \times \frac{8}{15} \times \frac{7}{14}=\frac{8}{\underline{15}} .

Initially there are (n + 16) discs and on the second selection there will be (n + 15) discs. Always consider all combinations as follows: 

\begin{aligned} \mathrm{P}(\text { one of each colour })=\frac{8}{21} & \Rightarrow 2 \times \frac{n}{n+16} \times \frac{16}{n+15}=\frac{8}{21} \\ & \Rightarrow 672 n=8(n+15)(n+16) \\ & \Rightarrow 84 n=(n+15)(n+16) \\ & \Rightarrow 84 n=n^2+31 n+240 \\ & \Rightarrow n^2-53 n+240=0 . \end{aligned}

n^2-53 n+240=0 \Rightarrow(n-48)(n-5)=0 \Rightarrow n=5 \text { or } 48

Probability And Quadratic Equations - More Examples

There is (3n + 1) discs in total for the first picking and so on the second picking there will be (3n) discs. 

\begin{aligned} \mathrm{P}(\mathrm{BB})=\frac{11}{24} & \Rightarrow \frac{2 n+1}{3 n+1} \times \frac{2 n}{3 n}=\frac{11}{24} \\ & \Rightarrow \frac{2 n+1}{3 n+1}=\frac{11}{16} \\ & \Rightarrow 16(2 n+1)=11(3 n+1) \\ & \Rightarrow 32 n+16=33 n+11 \\ & \Rightarrow n=5 \end{aligned}

This means there are 5 white and 11 black discs. 

\mathrm{P}(\mathrm{WW})=\frac{5}{16} \times \frac{4}{15}=\frac{1}{12}
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Probability And Quadratic Equations – 3 Final Examples

 
There are (2n + 3) discs in total. 
\begin{aligned} \mathrm{P}(\text { one of each colour })=\frac{28}{55} & \Rightarrow 2 \times \frac{n}{2 n+3} \times \frac{n+3}{2(n+1)}=\frac{28}{55} \\ & \Rightarrow 55 n(n+3)=28(2 n+3)(n+1) \\ & \Rightarrow 55 n^2+165 n=56 n^2+140 n+84 \\ & \Rightarrow n^2-25 n+84=0, \end{aligned}
 
 
n^2-25 n+84=0 \Rightarrow(n-4)(n-21)=0 \Rightarrow n=4 \text { or } n=21
 
So there is 4 red and 7 green OR 21 red and 24 green. 

P(R R)=\frac{21}{45} \times \frac{20}{44}=\frac{7}{33}

There are (x – 10) girls: 

\begin{aligned} \mathrm{P}(\text { one of each gender })=\frac{40}{77} & \Rightarrow 2 \times \frac{10}{x} \times \frac{x-10}{x-1}=\frac{40}{77} \\ & \Rightarrow 1540(x-10)=40 x(x-1) \\ & \Rightarrow 77(x-10)=2 x(x-1) \\ & \Rightarrow 77 x-770=2 x^2-2 x \\ & \Rightarrow 2 x^2-79 x+770=0, \end{aligned}

2 x^2-79 x+770=0 \Rightarrow(2 x-35)(x-22)=0 \Rightarrow x=22 \text { or } 17 \frac{1}{2}

So there are 22 students in the group. 

\begin{aligned} \mathrm{P}(\mathrm{CC})=\frac{1}{6} & \Rightarrow \frac{c}{c+5} \times \frac{c-1}{c+4}=\frac{1}{6} \\ & \Rightarrow 6 c(c-1)=(c+4)(c+5) \\ & \Rightarrow 6 c^2-6 c=c^2+9 c+20 \\ & \Rightarrow 5 c^2-15 c-20=0 \\ & \Rightarrow c^2-3 c-4=0 \\ & \Rightarrow(c-4)(c+1)=0 \\ & \Rightarrow c=4 \text { (since it must be a positive integer) } \end{aligned}

So there are 9 letters in the word. 

Consider all combinations, P(VC) and P(CV):

\mathrm{P}(\text { one of each })=2 \times \frac{4}{9} \times \frac{5}{8}=\frac{5}{9}

Conclusion:
 In conclusion, probability and quadratic equations play significant roles in mathematics and have practical applications in real-life scenarios. Understanding the probability of obtaining solutions to quadratic equations allows us to predict outcomes accurately. Moreover, the application of these concepts in real-life scenarios enables us to solve problems related to projectiles, probability events, and more. By mastering these topics, students can excel in the GCSE Higher Mathematics Paper and develop problem-solving skills that extend beyond the classroom.

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