If three numbers are randomly selected from the set \( \{1,2,3,\dots,50\} \), then the probability that they are in arithmetic progression is:
\( \frac{3}{25} \)
Step 1: Total Ways to Select Three Numbers
The given set is: \[ \{1,2,3,\dots,50\}. \] The number of ways to choose any three numbers from this set is: \[ \text{Total selections} = \binom{50}{3} = \frac{50!}{3!(50-3)!} = \frac{50 \times 49 \times 48}{6} = 19600. \]
Step 2: Counting Selections that Form an Arithmetic Progression
For three numbers to be in arithmetic progression, they must be of the form: \[ a-d, a, a+d. \] where \( a \) is the middle term, and \( d \) is the common difference. The constraints are: - \( a-d \geq 1 \), - \( a+d \leq 50 \). Rewriting: \[ 1 \leq a - d, \quad a + d \leq 50. \] The number of valid sequences is determined by choosing \( a \) and \( d \) such that they remain within the set boundaries. Counting possible values of \( (a,d) \): - The middle term \( a \) can be any value from 2 to 49. - The common difference \( d \) can range from 1 to \( \min(a-1, 50-a) \). Summing up all valid pairs: \[ 1 + 2 + \dots + 24 = \frac{24 \times 25}{2} = 300. \]
Step 3: Computing Probability
\[ P(\text{AP}) = \frac{\text{favorable cases}}{\text{total cases}} = \frac{300}{19600}. \] Simplifying: \[ P(\text{AP}) = \frac{3}{98}. \]
Step 4: Conclusion
Thus, the final probability that three randomly selected numbers form an arithmetic progression is: \[ \boxed{\frac{3}{98}}. \]
A student has to write the words ABILITY, PROBABILITY, FACILITY, MOBILITY. He wrote one word and erased all the letters in it except two consecutive letters. If 'LI' is left after erasing then the probability that the boy wrote the word PROBABILITY is: \
Given the vectors:
\[ \mathbf{a} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} \]
\[ \mathbf{b} = 3(\mathbf{i} - \mathbf{j} + \mathbf{k}) = 3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k} \]
where
\[ \mathbf{a} \times \mathbf{c} = \mathbf{b} \]
\[ \mathbf{a} \cdot \mathbf{x} = 3 \]
Find:
\[ \mathbf{a} \cdot (\mathbf{x} \times \mathbf{b} - \mathbf{c}) \]
A rectangle is formed by the lines \[ x = 4, \quad x = -2, \quad y = 5, \quad y = -2 \] and a circle is drawn through the vertices of this rectangle. The pole of the line \[ y + 2 = 0 \] with respect to this circle is:
The equation of a circle which passes through the points of intersection of the circles \[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \quad x^2 + y^2 + 4x + 2y + 1 = 0 \] and whose centre lies on the common chord of these circles is: