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14.Probability
medium
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$\frac{1}{2}$
B$\frac{3}{5}$
C$\frac{2}{3}$
D$\frac{4}{9}$
(JEE MAIN-2025)
Solution
$a=\text { number or dice } 1$
$b=\text { number on dice } 2$
$(a, b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1)$
Required probability
$=\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6}$
$=\frac{18}{36}=\frac{1}{2}$
$b=\text { number on dice } 2$
$(a, b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1)$
Required probability
$=\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6}$
$=\frac{18}{36}=\frac{1}{2}$
Standard 11
Mathematics
