Question

The sum of two positive numbers is 100. What is the probability that their product is greater than 1000?
A. $\dfrac{7}{9}$
B. $\dfrac{7}{{10}}$
C. $\dfrac{2}{5}$
D. None of these

Answer 1

Hint: First we will find the total number of pairs whose sum is 100. Then we will find the number of pairs whose product is greater than 1000. By using the probability formula, we will calculate the required probability.

Formula Used:
Probability of an event is $\dfrac{{{\rm{The}}\,\,{\rm{number}}\,\,{\rm{of}}\,\,{\rm{favorable}}\,\,{\rm{outcomes}}}}{{{\rm{The }}\,{\rm{number }}\,{\rm{of}}\,\,{\rm{total}}\,\,{\rm{number}}\,\,{\rm{outcomes}}}}$

Complete step by step solution:
Given that the sum of two positive numbers is 100.
The possible pairs whose sum is 100 are $\left( {1,99} \right),\,\left( {2,98} \right),\,\left( {3,97} \right),\, \cdots ,\left( {1,99} \right)$.
The number of pairs is 99.
Now we will find the number of pairs whose product is more than 1000.
$1 \times 99 = 99$
$2 \times 98 = 196$
$3 \times 97 = 291$
$4 \times 96 = 384$
$5 \times 95 = 475$
$6 \times 94 = 564$
$7 \times 93 = 651$
$8 \times 92 = 736$
$9 \times 91 = 819$
$10 \times 90 = 900$
$11 \times 89 = 979$
$12 \times 88 = 1056$
$13 \times 87 = 1131$
……
So on.
From the pair $\left( {12,88} \right)$ to $\left( {88,12} \right)$, the product of the pairs is more than 1000.
So, the number of pairs is $\left( {88 - 12 + 1} \right) = 77$.
Now we will apply the probability formula.
The number of favorable outcomes is 77.
The total number of outcomes is 99.
The probability is $\dfrac{{77}}{{99}}$.
Divide the denominator and numerator by 11.
$ = \dfrac{7}{9}$

Option ‘A’ is correct

Note: Students often make mistakes when calculating the total number of pairs. They consider the pairs $\left( {0,100} \right)$ and $\left( {100,0} \right)$. But 0 is not a positive number. Thus, they calculate the number of pairs is 101 which is incorrect.
But the correct number of pairs is 99.