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# A and B throw a die. The probability that A's throw is not greater than B's is :

$\begin {array} {1 1} (1)\;\large\frac{1}{2} & \quad (2)\;\large\frac{1}{6} \\ (3)\;\large\frac{7}{12} & \quad (4)\;\large\frac{5}{12} \end {array}$

Can you answer this question?

If B throws 1 then A can throw only 1.
If B throws 2 then A can throw only 1 and 2
$\therefore$ Required probability = $\large\frac{1}{6}.\large\frac{1}{6}+\large\frac{1}{6}.\large\frac{2}{6}+\large\frac{1}{6}.\large\frac{3}{6}+\large\frac{1}{6}.\large\frac{4}{6}+\large\frac{1}{6}.\large\frac{5}{6}+\large\frac{1}{6}.\large\frac{6}{6}$
$\large\frac{1}{36} [ 1+2+3+4+5+6]$
$\large\frac{1}{36} \times 21$
$= \large\frac{7}{12}$
Ans : (3)
answered Jan 3, 2014