A and B each throw a fair dice. The probability that A’s throw is not greater than B’s throw, is equal to
Answer & explanation
Correct answer: option 3
Let ‘A’s throw is r,r = 1, 2, 3, …6, then ‘B’s outcome can be (r, r + 1, ….6),
Thus required probability
$=\sum\limits_{r=1}^6 \frac{7-r}{36}=\sum\limits_{r=1}^6 \frac{r}{36}=\frac{7}{12}$