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Two dice are thrown, the probability that the sum of the points on the dice is 7 is

$\begin {array} {1 1} (A)\;\large\frac{5}{36} & \quad (B)\;\large\frac{6}{36} \\ (C)\;\large\frac{7}{36} & \quad (D)\;\large\frac{8}{36} \end {array}$

Can you answer this question?

The favourable cases are
$(1,6) (6,1)(2,5)(5,2)(3,4)(4,3)$
$\therefore$ Required probability = $\large\frac{6}{36}$
$=\large\frac{1}{6}$
Ans : (B)

answered Jan 6, 2014