A number is chosen at random from the set $\{1,2,3, \ldots , 2000\} $ Let $p$ be the probability that the chosen number is a multiple of $3$ or a multiple of $7$. Then the value of $500\, p$ is_____
Correct Answer: 214
Solution and Explanation
Step 1: Define the set E
We are given that $E$ is the set of numbers that are either multiples of 3 or multiples of 7. This means we are interested in counting the elements that are divisible by 3, divisible by 7, and avoiding double-counting those that are divisible by both 3 and 7 (i.e., divisible by 21).
Step 2: Set notation for n(E)
The number of elements in $E$ can be calculated using the principle of inclusion-exclusion:
$$ n(E) = \text{(multiples of 3)} + \text{(multiples of 7)} - \text{(multiples of 21)} $$
This formula is derived to avoid double-counting the multiples of both 3 and 7.
Step 3: Count the multiples of 3
The multiples of 3 form the sequence $3, 6, 9, \dots, 1998$. To find how many terms are in this sequence, we divide the largest multiple of 3 within the range (1998) by 3:
$$ \frac{1998}{3} = 666. $$
Thus, there are 666 multiples of 3.
Step 4: Count the multiples of 7
The multiples of 7 form the sequence $7, 14, 21, \dots, 1995$. To find how many terms are in this sequence, we divide the largest multiple of 7 within the range (1995) by 7:
$$ \frac{1995}{7} = 285. $$
Thus, there are 285 multiples of 7.
Step 5: Count the multiples of both 3 and 7 (multiples of 21)
The multiples of 21 (since $21 = 3 \times 7$) form the sequence $21, 42, 63, \dots, 1995$. To find how many terms are in this sequence, we divide the largest multiple of 21 within the range (1995) by 21:
$$ \frac{1995}{21} = 95. $$
Thus, there are 95 multiples of 21.
Step 6: Apply the inclusion-exclusion principle
Now, we can apply the inclusion-exclusion principle to find the total number of elements in $E$:
$$ n(E) = 666 + 285 - 95 = 856. $$
Thus, the total number of elements in $E$ is 856.
Step 7: Calculate the probability P(E)
The total number of possible outcomes (in this case, the total number of elements under consideration) is 2000. Therefore, the probability of selecting an element from $E$ is:
$$ P(E) = \frac{856}{2000}. $$
Step 8: Multiply the probability by 500
We are asked to find $P(E) \times 500$. Using the probability we calculated:
$$ P(E) \times 500 = \frac{856}{2000} \times 500 = \frac{856 \times 500}{2000} = \frac{856}{4} = 214. $$
Thus, the final result is 214.
Top Questions on Probability
- If the odds in favour of an event A are 3 : 4 and the odds against another independent event B are 7 : 4, find the probability that at least one of the events will happen.
- A random experiment has five outcomes \(w_1, w_2, w_3, w_4, w_5\). The probabilities of the occurrence of the outcomes \(w_1, w_2, w_4, w_5\) are respectively \( \frac{1}{6}, a, b, \frac{1}{12} \) such that \(12a + 12b - 1 = 0\). Then the probabilities of occurrence of the outcome \(w_3\) is:
- KCET - 2025
- Mathematics
- Probability
- Persons A and B decide to arrive and meet sometime between 7 and 8 pm. Whoever arrives first will wait for ten minutes for the other person. If the other person doesn't turn up inside ten minutes, then the person waiting will leave. What is the probability that they will meet?
- Let \( S \) be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set \( S \), one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is:
- JEE Main - 2025
- Mathematics
- Probability
- There are 20 girls and 15 boys in a class. 1/5 of the girls and 2/5 of the boys bring their lunch from home. One of the students who brings lunch from home is chosen at random. Find the probability that the student is a girl.
Questions Asked in JEE Advanced exam
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
- JEE Advanced - 2025
- Coordinate Geometry
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
- JEE Advanced - 2025
- Waves and Oscillations
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.