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Question

A natural number n is chosen from the first 50 natural numbers. What is the probability that \(n+\frac{50}{n}<50 \) ?

The correct answer is \(\frac{47}{50}\)

Understanding the Probability Question

The problem asks for the probability that a natural number \(n\), chosen from the first 50 natural numbers, satisfies a specific inequality. The first 50 natural numbers are 1, 2, 3, ..., 50.

The total number of possible outcomes when choosing a number from this set is 50. This forms our sample space.

The condition we need to check is \(n+\frac{50}{n}<50\).

Analyzing the Inequality \(n+\frac{50}{n}<50\)

To find the natural numbers \(n\) that satisfy this inequality, we can first rearrange it. Since \(n\) is a natural number, \(n > 0\). We can multiply both sides of the inequality by \(n\) without changing the direction of the inequality sign:

\[n \cdot \left(n + \frac{50}{n}\right) < 50 \cdot n\]

\[n^2 + 50 < 50n\]

Now, let's move all terms to one side to get a quadratic inequality:

\[n^2 - 50n + 50 < 0\]

Finding the Roots of the Related Quadratic Equation

To find the values of \(n\) for which the quadratic expression \(n^2 - 50n + 50\) is less than zero, we first find the roots of the corresponding quadratic equation \(n^2 - 50n + 50 = 0\). We can use the quadratic formula:

\[n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Here, \(a=1\), \(b=-50\), and \(c=50\).

\[n = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(1)(50)}}{2(1)}\]

\[n = \frac{50 \pm \sqrt{2500 - 200}}{2}\]

\[n = \frac{50 \pm \sqrt{2300}}{2}\]

We can simplify \(\sqrt{2300}\): \(\sqrt{2300} = \sqrt{100 \times 23} = 10\sqrt{23}\).

\[n = \frac{50 \pm 10\sqrt{23}}{2}\]

\[n = 25 \pm 5\sqrt{23}\]

Estimating the Roots and Finding Favorable Natural Numbers

Now, let's estimate the values of the roots. We know that \(4^2 = 16\) and \(5^2 = 25\), so \(\sqrt{23}\) is between 4 and 5, approximately 4.796.

  • Lower root: \(n_1 = 25 - 5\sqrt{23} \approx 25 - 5(4.796) = 25 - 23.98 = 1.02\)
  • Upper root: \(n_2 = 25 + 5\sqrt{23} \approx 25 + 5(4.796) = 25 + 23.98 = 48.98\)

The quadratic expression \(n^2 - 50n + 50\) represents an upward-opening parabola. The inequality \(n^2 - 50n + 50 < 0\) is satisfied for values of \(n\) that lie between the roots.

So, the inequality \(n+\frac{50}{n}<50\) is satisfied for approximately \(1.02 < n < 48.98\).

We are looking for natural numbers \(n\) from the set {1, 2, ..., 50} that fall within this range. The natural numbers greater than 1.02 are 2, 3, 4, ... and the natural numbers less than 48.98 are ..., 46, 47, 48.

The natural numbers \(n\) in the set {1, 2, ..., 50} that satisfy the inequality are {2, 3, 4, ..., 48}.

Let's count the number of such natural numbers. This is the number of favorable outcomes. The count is \(48 - 2 + 1 = 47\).

Total number of possible outcomes is 50 (since \(n\) is chosen from the first 50 natural numbers).

Calculating the Probability

The probability is the ratio of the number of favorable outcomes to the total number of outcomes.

\[\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\]

\[\text{Probability} = \frac{47}{50}\]

Summary of Results

The natural numbers \(n\) from 1 to 50 that satisfy the inequality \(n+\frac{50}{n}<50\) are {2, 3, ..., 48}. There are 47 such numbers. The total number of possible natural numbers is 50.

The probability is \(\frac{47}{50}\).

Revision Table: Probability Calculation Steps

Step Description Value/Expression
1 Identify the total possible outcomes (Sample Space) First 50 natural numbers (1 to 50), Total = 50
2 State the inequality condition \(n+\frac{50}{n}<50\)
3 Rearrange the inequality for \(n>0\) \(n^2 - 50n + 50 < 0\)
4 Find roots of the quadratic equation \(n^2 - 50n + 50 = 0\) \(n = 25 \pm 5\sqrt{23}\)
5 Estimate approximate range for \(n\) Approx. \(1.02 < n < 48.98\)
6 Identify natural numbers in the range (1 to 50) satisfying condition {2, 3, ..., 48}
7 Count favorable outcomes 47
8 Calculate Probability \(\frac{47}{50}\)

Additional Information: Quadratic Inequalities and Probability

Understanding how to solve quadratic inequalities is crucial for solving this type of probability problem. A quadratic inequality in the form \(ax^2 + bx + c < 0\) or \(ax^2 + bx + c > 0\) (or with \(\le\), \(\ge\)) can be solved by finding the roots of the corresponding equation \(ax^2 + bx + c = 0\).

  • If the parabola opens upwards (i.e., \(a > 0\)), the expression \(ax^2 + bx + c\) is negative (less than zero) between the roots and positive (greater than zero) outside the roots.
  • If the parabola opens downwards (i.e., \(a < 0\)), the expression is positive between the roots and negative outside the roots.

In this problem, the inequality is \(n^2 - 50n + 50 < 0\), and \(a=1\) (which is greater than 0). So the parabola opens upwards, and the inequality holds for values of \(n\) between the two roots we calculated.

Probability is defined as the likelihood of a specific event occurring. It is calculated as the ratio of the number of ways the event can occur (favorable outcomes) to the total number of possible outcomes in the sample space.

For this problem, the event is "the natural number \(n\) satisfies \(n+\frac{50}{n}<50\)". The sample space is the set of the first 50 natural numbers.

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Important Questions from Probability of Random Experiments

  1. A, B, C and D are mutually exclusive and exhaustive events.

    If 2P(A) = 3P(B) = 4P(C) = 5P(D), then what is 77P(A) equal to ?

  2. A fair coin is tossed 6 times. What is the probability of getting a result in the 6t h toss which is different from those obtained in the first five tosses ?

  3. Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is

  4. A biased coin with the probability of getting head equal to \(\frac{1}{4}\) is tossed five times. What is the probability of getting tail in all the first four tosses followed by head ? 

  5. Three dice are thrown. What is the probability that each face shows only multiples of 3 ?

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