A number x is chosen at random from the set {1, 2, 3, 4, ..., 100}. Define the event: A = the chosen number x satisfies $$\frac{(x-10)(x-50)}{(x-30)} \geq 0$$. Then P(A) is:

We are given a set of numbers from 1 to 100, and we choose a number $$x$$ at random. The event $$A$$ is defined by the inequality $$\frac{(x-10)(x-50)}{(x-30)} \geq 0$$. We need to find the probability $$P(A)$$, which is the number of integers $$x$$ in $$\{1, 2, \ldots, 100\}$$ that satisfy this inequality divided by 100.

First, note that the expression is undefined when the denominator is zero, i.e., at $$x = 30$$. Therefore, $$x = 30$$ cannot satisfy the inequality and must be excluded from the favorable outcomes.

The critical points where the numerator or denominator is zero are $$x = 10$$, $$x = 50$$, and $$x = 30$$. These points divide the real number line into intervals. We will test the sign of the expression in each interval to determine where the inequality holds. The intervals are:

• $$x < 10$$
• $$10 < x < 30$$
• $$30 < x < 50$$
• $$x > 50$$

We analyze each interval by picking a test point and evaluating the sign of each factor:

• For $$x < 10$$ (e.g., $$x = 5$$):
  • $$(x-10) = 5-10 = -5 < 0$$
  • $$(x-50) = 5-50 = -45 < 0$$
  • $$(x-30) = 5-30 = -25 < 0$$
  • Numerator: $$(-) \times (-) = +$$
  • Denominator: $$-$$
  • Fraction: $$\frac{+}{-} = -$$ (negative)
  Since the expression is negative, it does not satisfy $$\geq 0$$.
• For $$10 < x < 30$$ (e.g., $$x = 20$$):
  • $$(x-10) = 20-10 = 10 > 0$$
  • $$(x-50) = 20-50 = -30 < 0$$
  • $$(x-30) = 20-30 = -10 < 0$$
  • Numerator: $$(+) \times (-) = -$$
  • Denominator: $$-$$
  • Fraction: $$\frac{-}{-} = +$$ (positive)
  Since the expression is positive, it satisfies $$\geq 0$$.
• For $$30 < x < 50$$ (e.g., $$x = 40$$):
  • $$(x-10) = 40-10 = 30 > 0$$
  • $$(x-50) = 40-50 = -10 < 0$$
  • $$(x-30) = 40-30 = 10 > 0$$
  • Numerator: $$(+) \times (-) = -$$
  • Denominator: $$+$$
  • Fraction: $$\frac{-}{+} = -$$ (negative)
  Since the expression is negative, it does not satisfy $$\geq 0$$.
• For $$x > 50$$ (e.g., $$x = 60$$):
  • $$(x-10) = 60-10 = 50 > 0$$
  • $$(x-50) = 60-50 = 10 > 0$$
  • $$(x-30) = 60-30 = 30 > 0$$
  • Numerator: $$(+) \times (+) = +$$
  • Denominator: $$+$$
  • Fraction: $$\frac{+}{+} = +$$ (positive)
  Since the expression is positive, it satisfies $$\geq 0$$.

Next, we check the critical points:

• At $$x = 10$$: $$\frac{(10-10)(10-50)}{(10-30)} = \frac{0 \times (-40)}{-20} = \frac{0}{-20} = 0 \geq 0$$, so it satisfies.
• At $$x = 50$$: $$\frac{(50-10)(50-50)}{(50-30)} = \frac{40 \times 0}{20} = \frac{0}{20} = 0 \geq 0$$, so it satisfies.
• At $$x = 30$$: The expression is undefined, so it does not satisfy.

Combining the results, the inequality holds for:

• $$x = 10$$
• $$10 < x < 30$$
• $$x = 50$$
• $$x > 50$$

In interval notation, this is $$x \in [10, 30) \cup [50, \infty)$$. Since our set is $$\{1, 2, \ldots, 100\}$$, we consider integers from 1 to 100.

The favorable integers are:

• From 10 to 29 inclusive (since 30 is excluded): $$x = 10, 11, \ldots, 29$$.
• From 50 to 100 inclusive: $$x = 50, 51, \ldots, 100$$.

Now, we count these integers:

• Number of integers from 10 to 29: First term is 10, last term is 29, so the count is $$29 - 10 + 1 = 20$$.
• Number of integers from 50 to 100: First term is 50, last term is 100, so the count is $$100 - 50 + 1 = 51$$.

Total favorable outcomes = $$20 + 51 = 71$$.

Total possible outcomes = 100 (since the set has 100 integers).

Therefore, $$P(A) = \frac{71}{100} = 0.71$$.

Comparing with the options:

• A. 0.71
• B. 0.70
• C. 0.51
• D. 0.20

Hence, the correct answer is Option A.