Q.21. From a month of 31 days, 3 different dates are selected at random. If the probability that the

Q: What is the correct answer for JEE Main Mathematics Q21?

The correct answer is 944. Sol. Total number of ways to select 3 different dates from 31 days is ${31 \choose 3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 31 \times 5 \times 29 = 4495$. Let the three selected dates be $x, y, z$ such that $1 \le x 0$. So $x, x+d, x+2d$ are the dates. We have

Q: Q.21. From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to , where and , then is equal to ________.

The correct answer is 944. Sol. Total number of ways to select 3 different dates from 31 days is ${31 \choose 3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 31 \times 5 \times 29 = 4495$. Let the three selected dates be $x, y, z$ such that $1 \le x 0$. So $x, x+d, x+2d$ are the dates. We have

Question

Q.21. From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to , where  and , then  is equal to ________.

FirstInTest · Accepted Answer

Sol. Total number of ways to select 3 different dates from 31 days is ${31 \choose 3} = \frac{31 	imes 30 	imes 29}{3 	imes 2 	imes 1} = 31 	imes 5 	imes 29 = 4495$.
Let the three selected dates be $x, y, z$ such that $1 \le x < y < z \le 31$.
For these dates to be in an increasing A.P., let $y-x = d$ and $z-y = d$ for some common difference $d > 0$.
So $x, x+d, x+2d$ are the dates.
We have $1 \le x < x+d < x+2d \le 31$.
This implies $x \ge 1$ and $x+2d \le 31$.
Also, $x, x+d, x+2d$ must have the same parity (all odd or all even) for $d$ to be an integer. No, this is incorrect. If $d$ is an integer, then $x$ and $x+2d$ have the same parity. $x+d$ has opposite parity to $x$ if $d$ is odd, and same parity as $x$ if $d$ is even.
However, for $x, y, z$ to be in A.P., $x+z = 2y$. This means $x$ and $z$ must have the same parity.
So, either $x$ and $z$ are both odd, or $x$ and $z$ are both even.
Case 1: $x$ and $z$ are both odd.
There are 16 odd numbers between 1 and 31 (1, 3, ..., 31).
If we choose any two distinct odd numbers, say $x$ and $z$, then $y = (x+z)/2$ will be an integer. Since $x$ and $z$ are odd, $x+z$ is even, so $y$ is an integer. Also, $x < y < z$.
So, the number of ways to choose two distinct odd numbers from 16 odd numbers is ${16 \choose 2} = \frac{16 	imes 15}{2} = 8 	imes 15 = 120$.
Case 2: $x$ and $z$ are both even.
There are 15 even numbers between 1 and 31 (2, 4, ..., 30).
If we choose any two distinct even numbers, say $x$ and $z$, then $y = (x+z)/2$ will be an integer. Since $x$ and $z$ are even, $x+z$ is even, so $y$ is an integer. Also, $x < y < z$.
So, the number of ways to choose two distinct even numbers from 15 even numbers is ${15 \choose 2} = \frac{15 	imes 14}{2} = 15 	imes 7 = 105$.
Total number of A.P. triplets = $120 + 105 = 225$.
Probability $P = \frac{225}{4495}$.
To simplify the fraction, divide by 5:
$P = \frac{45}{899}$.
We need to check if $\gcd(45, 899) = 1$.
$45 = 3^2 	imes 5$.
$899 = 29 	imes 31$. (Since $30^2 = 900$, check primes around 30).
So $\gcd(45, 899) = 1$.
Thus $a=45$ and $b=899$.
$a+b = 45 + 899 = 944$.

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