Question

The sum of two positive numbers is $100$. The probability that their product is greater than $1000$ is

• A. $\frac{7}{9}$
• B. $\frac{7}{10}$
• C. $\frac{2}{5}$
• D. None of the above.

Answer 1

The correct option is A

$\frac{7}{9}$

Explanation for the correct option:

Step 1: Calculate the number of possible outcomes

Let $x$ and $y$ be the two numbers.

The formula to find the probability of an event is given as,

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

where $P\left(A\right)$ is the probability of the desired outcome, $n\left(A\right)$ is the number of desired outcome and $n\left(S\right)$ is the total possible outcomes.

Given, $x+y=100$

The possible values of $x$ and $y$ represented as $\left(x,y\right)$ (i.e., the sample space) are

$$\begin{gathered} \left\{\left(1,99\right),\left(2,98\right),\left(3,97\right),\dots ,\left(97,3\right),\left(98,2\right),\left(99,1\right)\right\} \\ \therefore n\left(S\right)=99 \end{gathered}$$

[Note: $x$ or $y$ cannot be $0$ because $0$ is not a positive integer]

Step 2: Calculate the number of desired outcomes

We have to find the probability that $xy>1000$.

$\begin{array}{rcl}& & \begin{array}{ccc}\Rightarrow & x>\frac{1000}{y}& \\ \Rightarrow & x>\frac{1000}{100-x}& \left[\because x+y=100\Rightarrow y=100-x\right]\\ \Rightarrow & 100x-x^2>1000& \\ \Rightarrow & 100x-x^2-1000>0& \\ \Rightarrow & x^2-100x+1000<0& \end{array}\end{array}$

By using the formula to find the roots of the quadratic equation,

$\begin{array}{ccc}& x<\frac{-\left(-100\right)\pm \sqrt{{\left(-100\right)}^2-4\times 1\times 1000}}{2}& \\ \Rightarrow & x<\frac{100\pm \sqrt{6000}}{2}& \\ \Rightarrow & x<\frac{100\pm \sqrt{4\times 15\times 100}}{2}& \\ \Rightarrow & x<\frac{100\pm 20\sqrt{15}}{2}& \\ \Rightarrow & x<\frac{100+20\sqrt{15}}{2},& x>\frac{100-20\sqrt{15}}{2}\\ \Rightarrow & x<88.73,& x>11.27\end{array}$

[Note: the symbols$>,<,=$ are called relational symbols and they can be substituted in place of each other and will denote their respective meanings. So, the use of $<$ in the quadratic equation is justified]

Thus, when $x$ is between $11.27$ and $88.73$, it is true that $xy>1000$.

Thus, the possible values of $x$ are $12,13,14,\dots ,86,87,88$.

We can observe that the above sequence is in AP. The $N^{\mathrm{th}}$ term of an AP is defined as $a_N=a+(N-1)d$, where $a$ is the first term and $d$ is the common difference.

Here, $a_N=88$, $a=12$ and $d=1$.

Thus,

$\begin{array}{cc}& 88=12+(N-1)1\\ \Rightarrow & N-1=76\\ \Rightarrow & N=77\end{array}$

Therefore, $n\left(A\right)=77$

Step 3: Calculate the desired probability

$\begin{array}{rcl}\therefore P\left(A\right)& =& \frac{77}{99}\\ & =& \frac{7}{9}\end{array}$

Hence, $\mathrm{Option}\left(\mathrm{A}\right)$is correct.