Search any question & find its solution
Question:
Answered & Verified by Expert
A die has two faces each with number ' 1 ', three faces each with number ' 2 ' and one face with number ' 3 '. If die is rolled once, determine
(i) $P(2)$
(ii) $P(1$ or 3$)$
(iii) $P($ not 3$)$
(i) $P(2)$
(ii) $P(1$ or 3$)$
(iii) $P($ not 3$)$
Solution:
1369 Upvotes
Verified Answer
Total number of exhaustive cases $=6$
(i) 3 faces are marked with 2.
Number of favourable cases $=3$
Required probability $=P(2)=\frac{3}{6}=\frac{1}{2}$
(ii) 2 faces are marked $1, P(1)=\frac{2}{6}$
1 face is marked with $3, P(3)=\frac{1}{6}$
$P(1$ or 3$)=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
(iii) Only 1 face is marked with 3 .
5 faces are marked with number which is not 3
$\therefore$ No. of favourable cases $=5$
$\therefore P($ not 3$)=\frac{5}{6}$
(i) 3 faces are marked with 2.
Number of favourable cases $=3$
Required probability $=P(2)=\frac{3}{6}=\frac{1}{2}$
(ii) 2 faces are marked $1, P(1)=\frac{2}{6}$
1 face is marked with $3, P(3)=\frac{1}{6}$
$P(1$ or 3$)=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
(iii) Only 1 face is marked with 3 .
5 faces are marked with number which is not 3
$\therefore$ No. of favourable cases $=5$
$\therefore P($ not 3$)=\frac{5}{6}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.