From a month of 31 days, 3 different dates are selected at random.

Question

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 $\frac{a}{b}$, where $a, b \in$ N and $\operatorname{gcd}(a, b)=1$, then $a+b$

Accepted Answer

Total number of ways to select any $3$ different dates from $31$ days is

$(\frac{31}{3})$

Now we count the number of selections in which the three dates are in an increasing A.P.

Let the three dates be

$a-d,a,a+d$

where $d\ge 1$ is an integer.

So we need all triples of the form

$x,x+r,x+2r$

with

$x\ge 1,\, x+2r\le 31$

For a fixed common difference $r$, the first term $x$ can be chosen from

$1,2,3,\ldots ,31-2r$

So number of A.P.s with common difference $r$ is

$31-2r$

Now $r$ can take values

$1,2,3,\ldots ,15$

because

$x+2r\le 31\Rightarrow 2r\le 30\Rightarrow r\le 15$

Hence total favourable selections are

$\sum _{15}^{r=1}(31-2r)$

This is

$15\cdot 31-2\sum _{15}^{r=1}r$

$=465-2\cdot \frac{15\cdot 16}{2}$

$=465-240$

$=225$

Therefore,

$P=\frac{225}{(\frac{31}{3})}$

Now,

$(\frac{31}{3})=\frac{31\cdot 30\cdot 29}{6}=4495$

So,

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

Divide numerator and denominator by $5$:

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

Thus,

$a=45,\, b=899$

Therefore,

$a+b=45+899=944$

So the required answer is

$944$

Solutions · 1