Probability of Selecting Three Dates in Increasing AP from a 31 Day Month

To find the probability that three randomly selected dates from a 31-day month are in an increasing arithmetic progression (A.P.), we analyze the total number of outcomes and the number of favorable outcomes.

Step 1: Calculate the total number of ways to select 3 dates

The total number of ways to choose $3$ distinct dates out of $31$ days is given by the combination formula $\binom{n}{r}$:

$N_{\text{total}} = \binom{31}{3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 31 \times 5 \times 29 = 4495$

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Step 2: Calculate the number of favorable ways to form an A.P.

Let the three chosen dates arranged in increasing order be $x, y, z$ such that $1 \le x < y < z \le 31$.

For these three dates to form an increasing arithmetic progression, the middle term $y$ must be the arithmetic mean of $x$ and $z$:

$y = \frac{x + z}{2} \implies x + z = 2y$

This condition requires that $x$ and $z$ must have the same parity (both must be odd or both must be even) so that their sum $x + z$ is even and $y$ is an integer. Once $x$ and $z$ of the same parity are chosen, the middle element $y$ is uniquely determined.

In the set of dates $\{1, 2, 3, \dots, 31\}$:

• The number of odd dates is $16$ (i.e., $1, 3, 5, \dots, 31$).
• The number of even dates is $15$ (i.e., $2, 4, 6, \dots, 30$).

The number of ways to choose two dates of the same parity (which uniquely determine the triplet $(x, y, z)$) is:

• Choosing 2 odd dates: $\binom{16}{2} = \frac{16 \times 15}{2} = 120$
• Choosing 2 even dates: $\binom{15}{2} = \frac{15 \times 14}{2} = 105$

Thus, the total number of favorable triplets is:

$N_{\text{favorable}} = 120 + 105 = 225$

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Step 3: Compute the probability and find $a + b$

The probability $P$ is given by:

$P = \frac{N_{\text{favorable}}}{N_{\text{total}}} = \frac{225}{4495}$

Dividing both the numerator and denominator by their greatest common divisor $5$:

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

Since $899 = 29 \times 31$ and $45 = 3^2 \times 5$, we have $\text{gcd}(45, 899) = 1$.

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

Finally, we compute $a + b$:

$a + b = 45 + 899 = 944$