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
Solution
<p>$\mathrm{a}=$ number or dice 1</p>
<p>$\mathrm{b}=$ number on dice 2</p>
<p>$(a, b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1)$</p>
<p>Required probability</p>
<p>$$\begin{aligned}
& =\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6} \\
& =\frac{18}{36}=\frac{1}{2}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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