Probability 3 Question 6
6. Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E _1$ is the event that die $A$ shows up four, $E _2$ is the event that die $B$ shows up two and $E _3$ is the event that the sum of numbers on both dice is odd, then which of the following statements is not true?
(2016 Main) (a) $E _1$ and $E _2$ are independent
(b) $E _2$ and $E _3$ are independent
(c) $E _1$ and $E _3$ are independent
(d) $E _1, E _2$ and $E _3$ are independent
Show Answer
Solution:
- Clearly, $E _1={(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}$
and $\quad \begin{aligned} E _2 & ={(1,2),(2,2),(3,2),(4,2),(5,2),(6,2)} \ E _3 & ={(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),\end{aligned}$
$(3,2),(3,4),(3,6),(4,1),(4,3),(4,5)$, $(5,2),(5,4),(5,6),(6,1),(6,3),(6,5)}$
$\Rightarrow \quad P\left(E _1\right)=\frac{6}{36}=\frac{1}{6}, P\left(E _2\right)=\frac{6}{36}=\frac{1}{6}$
and $P\left(E _3\right)=\frac{18}{36}=\frac{1}{2}$
Now, $P\left(E _1 \cap E _2\right)=P$ (getting 4 on die $A$ and 2 on die $B$ )
$$ =\frac{1}{36}=P\left(E _1\right) \cdot P\left(E _2\right) $$
$P\left(E _2 \cap E _3\right)=P$ (getting 2 on die $B$ and sum of numbers on both dice is odd)
$$ =\frac{3}{36}=P\left(E _2\right) \cdot P\left(E _3\right) $$
$P\left(E _1 \cap E _3\right)=P$ (getting 4 on die $A$ and sum of numbers on both dice is odd)
$$ =\frac{3}{36}=P\left(E _1\right) \cdot P\left(E _3\right) $$
and $P\left(E _1 \cap E _2 \cap E _3\right)=P$ [getting 4 on die $A, 2$ on die $B$ and sum of numbers is odd]
$$ =P(\text { impossible event })=0 $$
Hence, $E _1, E _2$ and $E _3$ are not independent.
