JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Probability

Each die shows one of $1,2,3,5,7,11$, so there are $6\times 6=36$ equally likely ordered outcomes. Counting the admissible second value for each first value: with $1$ the second may be $1,2,3,5$ or $7$, five ways; with $2$ it may be $1,2,3$ or $5$, four ways; with $3$ it may be $1,2,3$ or $5$, four ways; with $5$ it may be $1,2$ or $3$, three ways; with $7$ only $1$, one way; and with $11$ no value works. That gives $5+4+4+3+1=17$ favourable outcomes, so the probability that the sum is at most $8$ is $\frac{17}{36}$.