ProbabilitymediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Probability 3 Dates in AP: a + b = 944 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

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 ab\dfrac{a}{b}, where a,b∈Na,b\in\mathbb{N} and gcd⁡(a,b)=1\gcd(a,b)=1, then a+ba+b is equal to
Solution
Answer: 944 (± 0.01)
Step 1:
(313)=31⋅30⋅296=4495.\binom{31}{3}=\frac{31\cdot30\cdot29}{6}=4495.
 Step 2: For A.P. a<b<ca<b<c, b=a+c2⇒a,cb=\dfrac{a+c}2\Rightarrow a,c same parity. Among 1,…,311,\ldots,31: 1616 odd, 1515 even:
favourable=(162)+(152)=16⋅152+15⋅142=120+105=225.\text{favourable}=\binom{16}{2}+\binom{15}{2}=\frac{16\cdot15}2+\frac{15\cdot14}2=120+105=225.
 Step 3:
P=2254495=45899=ab.P=\frac{225}{4495}=\frac{45}{899}=\frac ab.
 Step 4: a=45, b=899a=45,\ b=899:
a+b=45+899=944.a+b=45+899=944.
 Correct answer: 944
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