One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is -

a=number or dice ...1 b= number on dice ...2 (a, b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1) Required probability =(2 / 6) x (2 / 6)+(1 / 6) x (1 / 6)+(2 / 6) x (2 / 6)+(2 / 6) x (2 / 6)+(1 / 6) x (2 / 6)+(2 / 6) x (1 / 6)+(1 / 6) x (2 / 6) =(18 / 36)=(1 / 2)

One die has two faces marked 1 two faces marked 2 one f... | (Mathematics)

a=number or dice } ...1 $\\$ b= number on dice } ...2 $\\$ (a, b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1)

$\\$ Required probability

$\\=\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6} \\$

$=\frac{18}{36}=\frac{1}{2}$

Explanation

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