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14.Probability
normal
Let two fair dices $A$ and $B$ are thrown. Then the probability that number appears on dice $A$ is greater than number appears on dice $B$ is
A
$\frac {5}{12}$
B
$\frac {1}{2}$
C
$\frac {3}{4}$
D
$\frac {7}{17}$
Solution
Sample space $= 36$
$A>B$ then
$(6,1)(6,2)(6,3)(6,4)(6,5)$
$(5,1)(5,2)(5,3)(5,4)$
$(4,1)(4,2)(4,3)(3,1)(3,2) $ and $(2,1)$
Probability $=\frac{15}{36}=\frac{5}{12}$
Standard 11
Mathematics
