Two smallest squares are chosen one by one on a chessboard. The probability that they have a side in common is:
Show Hint
- \(\frac{1}{9}\)
- \(\frac{2}{7}\)
- \(\frac{1}{18}\)
- \(\frac{5}{18}\)
The Correct Option is C
Solution and Explanation
Step 1: The problem asks for the probability that two randomly chosen smallest squares on a chessboard share a side. A standard chessboard consists of \(8 \times 8\) squares, which means there are 64 total small squares.
Step 2: The number of possible pairs of squares is given by \(\binom{64}{2}\).
Step 3: The number of favorable outcomes where two squares have a side in common is based on the number of adjacent squares. Each square has up to 4 adjacent squares (except for the edge squares), and the total number of such adjacent square pairs is fewer than 64, as we are restricted to squares that share a side.
Step 4: By counting the adjacent pairs (both horizontally and vertically) and dividing by the total number of pairs, we obtain the probability \(\frac{1}{18}\).
Top Questions on Probability
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let \( A_1 \): People with good health,
\( A_2 \): People with average health,
and \( A_3 \): People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category \( A_1, A_2 \) and \( A_3 \) are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions:
(i) A person was tested randomly. What is the probability that he/she has contracted the disease?}
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category \( A_2 \)?- If A and B are two events such that $P(B) = \frac{1}{5}$, $P(A | B) = \frac{2}{3}$ and $P(A \cup B) = \frac{3}{5}$, then $P(A)$ is :
- If \( P(A) = \frac{1}{5}, P(B) = \frac{3}{5} \) and \( P(A \cap B) = \frac{2}{5} \), then \( P(A' \cup B') \) is :
- Bag I contains 4 white and 5 black balls. Bag II contains 6 white and 7 black balls. A ball drawn randomly from Bag I is transferred to Bag II and then a ball is drawn randomly from Bag II. Find the probability that the ball drawn is white.
- A meeting will be held only if all three members A, B and C are present. The probability that member A does not turn up is 0.10, member B does not turn up is 0.20 and member C does not turn up is 0.05. The probability of the meeting being cancelled is:
Questions Asked in WBJEE exam
- Which logic gate is represented by the following combination of logic gates?
- WBJEE - 2025
- Logic gates
- If \( {}^9P_3 + 5 \cdot {}^9P_4 = {}^{10}P_r \), then the value of \( 'r' \) is:
- WBJEE - 2025
- permutations and combinations
- If \( 0 \leq a, b \leq 3 \) and the equation \( x^2 + 4 + 3\cos(ax + b) = 2x \) has real solutions, then the value of \( (a + b) \) is:
- WBJEE - 2025
- Trigonometric Equations
- Let \( f(\theta) = \begin{vmatrix} 1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \end{vmatrix} \). Suppose \( A \) and \( B \) are respectively the maximum and minimum values of \( f(\theta) \). Then \( (A, B) \) is equal to:
- If \( a, b, c \) are in A.P. and if the equations \( (b - c)x^2 + (c - a)x + (a - b) = 0 \) and \( 2(c + a)x^2 + (b + c)x = 0 \) have a common root, then
- WBJEE - 2025
- Quadratic Equations