Question

Two dice are thrown simultaneously. What is the probability of obtaining a multiple of 2 on one of them and a multiple of 3 on the other

• A. $\frac { 5 } { 36 }$
• B. $\frac { 11 } { 36 }$
• C. $\frac { 1 } { 6 }$
• D. $\frac { 1 } { 3 }$

Answer 1

Answer: (B)

$\frac { 11 } { 36 }$

Answer 2

Favourable cases for one are three i.e. 2, 4 and 6 and for other are two i.e. 3, 6.

Hence required probability $= \left[ \left( \frac { 3 \times 2 } { 36 } \right) 2 - \frac { 1 } { 36 } \right] = \frac { 11 } { 36 }$

{As same way happen when dice changes numbers among themselves}