Question

In an experimental performance of a single throw of a pair of unbiased normal dice, let three events $E_1,E_2$ and $E_3$ are defined as follows:
$E_1$ : getting prime numbered face on each dice.
$E_2$ : getting the same number on each dice.
$E_3$ : getting total on two dice equal to 4 .
Which of the following is/are TRUE?

• A. The probabilities $P\left(E_1\right),P\left(E_2\right),P\left(E_3\right)$ are in A.P.
• B. The events $E_1$ and $E_2$ are independent.
• C. $P\left(E_3/E_1\right)=\frac{2}{9}$.
• D. $P\left(E_1+E_2\right)+P\left(E_2-E_3\right)=\frac{17}{36}$

Answer 1

Answer: (A), (D)

$E_1=\{(2,2),(2,3),(3,2),(2,5),(5,2),(3,5),(5,3),(3,3),(5,5)\}$
$E_2=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$
$E_3=\{(1,3),(3,1),(2,2)\}$
Now $P\left(E_1\right)=\frac{9}{36};P\left(E_2\right)=\frac{6}{36};P\left(E_3\right)=\frac{3}{36}$
Clearly $P\left(E_1\right),P\left(E_2\right),P\left(E_3\right)$ are in A.P. $\Rightarrow (A)$ is correct.
$P\left(E_1\cap E_2\right)=\frac{3}{36}=\frac{1}{12}\ne P\left(E_1\right)P\left(E_2\right)\Rightarrow$ (B) is incorrect.
Now $P\left(E_3/E_1\right)=\frac{P\left(E_3\cap E_1\right)}{P\left(E_1\right)}=\frac{1}{9}\Rightarrow (C)$ is incorrect.
Also $P\left(E_1+E_2\right)+P\left(E_2-E_3\right)=\left[P\left(E_1\right)+P\left(E_2\right)-P\left(E_1\cap E_2\right)\right]+[P\left(E_2\right)-P\left(E_2\cap E_3\right)$
$=\frac{9}{36}+\frac{12}{36}-\frac{3}{36}-\frac{1}{36}=\frac{17}{36}\Rightarrow$
(D) is correct.