JEE Main 2026 — Probability Question with Solution

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

${}^{31}{C}_3=\frac{31\times 30\times 29}{3\times 2\times 1}=4495$

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

$2y=x+z$

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

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

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

${}^{16}{C}_2=\frac{16\times 15}{2}=120$

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

${}^{15}{C}_2=\frac{15\times 14}{2}=105$

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

$120+105=225$

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

$P=\frac{225}{4495}$

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

$P=\frac{45}{899}$

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

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

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

Answer: $944$