JEE Main 2026 — Mathematics Probability & Statistics

The total number of ways to select $3$3 different dates from a month of $31$31 days is given by choosing $3$3 days out of $31$31:

$$^{31}C_{3} = \dfrac{31 \times 30 \times 29}{3 \times 2 \times 1} = 4495$$31C3​=3×2×131×30×29​=4495

Let the three selected dates be $x, y, z$x,y,z such that $x < y < z$x<y<z. For these dates to form an increasing Arithmetic Progression (A.P.), they must satisfy the condition:

$$2y = x + z$$2y=x+z

For $y$y to be an integer, the sum $x + z$x+z must be an even number. This is only possible if both $x$x and $z$z are either odd or even. Once $x$x and $z$z are chosen, $y$y is uniquely determined as their exact average.

In a month of $31$31 days, the number of odd dates ($1, 3, 5, \dots, 31$1,3,5,…,31) is $16$16, and the number of even dates ($2, 4, 6, \dots, 30$2,4,6,…,30) is $15$15.

The number of ways to choose $x$x and $z$z such that both are odd is:

$$^{16}C_{2} = \dfrac{16 \times 15}{2} = 120$$16C2​=216×15​=120

The number of ways to choose $x$x and $z$z such that both are even is:

$$^{15}C_{2} = \dfrac{15 \times 14}{2} = 105$$15C2​=215×14​=105

The total number of favorable selections that form an increasing A.P. is:

$$120 + 105 = 225$$120+105=225

The probability that the selected dates are in an increasing A.P. is:

$$P = \dfrac{225}{4495}$$P=4495225​

Dividing the numerator and the denominator by $5$5, we get:

$$P = \dfrac{45}{899}$$P=89945​

Since $45 = 3^2 \times 5$45=32×5 and $899 = 29 \times 31$899=29×31, they share no common factors, meaning $\gcd(45, 899) = 1$gcd(45,899)=1.

Thus, $a = 45$a=45 and $b = 899$b=899.

Finally, $a + b = 45 + 899 = 944$a+b=45+899=944.

Answer: $944$944