Browse Questions

A and B throw a dice each. The probability that A's throw is not greater than B's throw is

$\begin {array} {1 1} (A)\;\large\frac{7}{12} & \quad (B)\;\large\frac{5}{12} \\ (C)\;\large\frac{1}{6} & \quad (D)\;\large\frac{1}{2} \end {array}$

When A gets 1, B can get any of the numbers 1,2,3,4,5,6.
When A gets 2, B can get any of the numbers 2,3,4,5,6.
Favourable number of ways = 6+5+.....+1=21
Probability that A's throw is not greater than B's throw is $\large\frac{21}{36}$
= $\large\frac{7}{21}$
Hence (A) is the correct option.