KCET · Maths · Probability
Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 the probability of getting a sum as 3 is
- A \(\frac{1}{18}\)
- B \(\frac{5}{18}\)
- C \(\frac{1}{5}\)
- D \(\frac{2}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{5}\)
Step-by-step Solution
Detailed explanation
Let \(E_{A}=\) The event that the sum of numbers on the dice was less than 6 .
\(\begin{aligned}
&E_{A}=\{(1,4),(4,1),(2,3),(3,2),(2,2),(1,3),(3, \\
&1),(1,2),(2,1),(1,1)\} \\
&\Rightarrow \quad n\left(E_{A}\right)=10
\end{aligned}\)
\(E_{B}=\) The event that the sum of numbers on the dice is 3
\(\begin{aligned}
&\quad E_{B}=\{(1,2),(2,1)\} \\
&\Rightarrow \quad n\left(E_{2}\right)=2 \\
&\therefore \text { Required probability }=\frac{2}{10}=\frac{1}{5}
\end{aligned}\)
\(\begin{aligned}
&E_{A}=\{(1,4),(4,1),(2,3),(3,2),(2,2),(1,3),(3, \\
&1),(1,2),(2,1),(1,1)\} \\
&\Rightarrow \quad n\left(E_{A}\right)=10
\end{aligned}\)
\(E_{B}=\) The event that the sum of numbers on the dice is 3
\(\begin{aligned}
&\quad E_{B}=\{(1,2),(2,1)\} \\
&\Rightarrow \quad n\left(E_{2}\right)=2 \\
&\therefore \text { Required probability }=\frac{2}{10}=\frac{1}{5}
\end{aligned}\)
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