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Probability question

2023 · Shift 2 · Q34
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Probability question

2023 · Shift 2 · Q34

JEE AdvancedMathematicsProbabilityNumerical+3 / −1
Consider the 6×66 \times 66×6 square in the figure. Let A1,A2,…,A49A_1, A_2, \ldots, A_{49}A1​,A2​,…,A49​ be the points of intersections (dots in the picture) in some order. We say that AiA_iAi​ and AjA_jAj​ are friends if they are adjacent along a row or along a column. Assume that each point AiA_iAi​ has an equal chance of being chosen. JEE Advanced 2023 Paper 2 Online Mathematics - Probability Question 6 English ComprehensionTwo distinct points are chosen randomly out of the points A1,A2,…,A49A_1, A_2, \ldots, A_{49}A1​,A2​,…,A49​. Let ppp be the probability that they are friends. Then the value of 7p7 p7p is :
Numerical answer
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Correct answer: 0.50

1. Understanding the Grid Setup

The problem describes a 6×66 \times 66×6 square grid, which is composed of 6×6=366 \times 6 = 366×6=36 small squares. The points of intersection, labeled A1,A2,…,A49A_1, A_2, \ldots, A_{49}A1​,A2​,…,A49​, are the vertices of these small squares. A grid of n×nn \times nn×n squares has (n+1)×(n+1)(n+1) \times (n+1)(n+1)×(n+1) intersection points. For a 6×66 \times 66×6 grid, this means we have (6+1)×(6+1)=7×7=49(6+1) \times (6+1) = 7 \times 7 = 49(6+1)×(6+1)=7×7=49 points. These points are arranged in a 7×77 \times 77×7 lattice.

2. Calculating the Total Number of Outcomes

We are choosing two distinct points randomly from the 49 available points. The total number of ways to select two distinct points is the number of combinations of 49 items taken 2 at a time, denoted as (492)\binom{49}{2}(249​).

Total outcomes=(492)=49!2!(49−2)!=49×482=49×24=1176\text{Total outcomes} = \binom{49}{2} = \frac{49!}{2!(49-2)!} = \frac{49 \times 48}{2} = 49 \times 24 = 1176Total outcomes=(249​)=2!(49−2)!49!​=249×48​=49×24=1176

3. Counting the Favorable Outcomes

A favorable outcome is a pair of points that are "friends". Two points are friends if they are adjacent along a row or a column. We can count the total number of such pairs by counting the horizontal and vertical adjacent pairs separately.

  • Horizontal Pairs: The grid has 7 rows. In each row, there are 7 points. The number of adjacent pairs in a single row is 7−1=67 - 1 = 67−1=6. For example, in the first row, the pairs are (A1,1,A1,2),(A1,2,A1,3),…,(A1,6,A1,7)(A_{1,1}, A_{1,2}), (A_{1,2}, A_{1,3}), \ldots, (A_{1,6}, A_{1,7})(A1,1​,A1,2​),(A1,2​,A1,3​),…,(A1,6​,A1,7​). Since there are 7 rows, the total number of horizontal friendly pairs is: Nhorizontal=7 rows×6 pairs/row=42N_{\text{horizontal}} = 7 \text{ rows} \times 6 \text{ pairs/row} = 42Nhorizontal​=7 rows×6 pairs/row=42

  • Vertical Pairs: Similarly, the grid has 7 columns. In each column, there are 7 points. The number of adjacent pairs in a single column is 7−1=67 - 1 = 67−1=6. Since there are 7 columns, the total number of vertical friendly pairs is: Nvertical=7 columns×6 pairs/column=42N_{\text{vertical}} = 7 \text{ columns} \times 6 \text{ pairs/column} = 42Nvertical​=7 columns×6 pairs/column=42

The total number of favorable outcomes is the sum of the number of horizontal and vertical pairs, as these two sets of pairs are disjoint.

Favorable outcomes=Nhorizontal+Nvertical=42+42=84\text{Favorable outcomes} = N_{\text{horizontal}} + N_{\text{vertical}} = 42 + 42 = 84Favorable outcomes=Nhorizontal​+Nvertical​=42+42=84

4. Calculating the Probability, ppp

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

p=Favorable outcomesTotal outcomes=841176p = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{84}{1176}p=Total outcomesFavorable outcomes​=117684​

To simplify the fraction:

p=8449×24=12×77×7×24=127×24=17×2=114p = \frac{84}{49 \times 24} = \frac{12 \times 7}{7 \times 7 \times 24} = \frac{12}{7 \times 24} = \frac{1}{7 \times 2} = \frac{1}{14}p=49×2484​=7×7×2412×7​=7×2412​=7×21​=141​

5. Calculating the Final Value

The question asks for the value of 7p7p7p.

7p=7×p=7×114=714=12=0.57p = 7 \times p = 7 \times \frac{1}{14} = \frac{7}{14} = \frac{1}{2} = 0.57p=7×p=7×141​=147​=21​=0.5

Thus, the value of 7p7p7p is 0.5.

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