Question

Two dice are thrown, 1 is the event that the sum of their

Answer 1

Given:

Two dice are thrown.

$E_1$ is the event that the sum of their dots is a prime number

$E_2$ is the event that 5 is the dot on the top of second die.

To find:

Whether $P(E_1\cap E_2)=P(E_1)\cdot P(E_2)$ is true or false.

Solution:

If two dice thrown, then the total possible outcomes are:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).

$E_1$ is the event that the sum of their dots is a prime number.

$E_1=\{(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)\}$

$P(E_1)=\dfrac{15}{36}$

$P(E_1)=\dfrac{5}{12}$

$E_2$ is the event that 5 is the dot on the top of second die.

$E_2=\{(1,5), (2,5),(3,5),(4,5),(5,5),(6,5)\}$

$P(E_2)=\dfrac{6}{36}$

$P(E_2)=\dfrac{1}{6}$

The intersection of these two events is:

$E_1\cap E_2=\{(2,5),(6,5)\}$

$P(E_1\cap E_2)=\dfrac{2}{36}$

$P(E_1\cap E_2)=\dfrac{1}{18}$

Now,

$P(E_1)\cdot P(E_2)=\dfrac{5}{12}\cdot \dfrac{1}{6}$

$P(E_1)\cdot P(E_2)=\dfrac{5}{72}$

$P(E_1)\cdot P(E_2)\neq P(E_1\cap E_2)$

Therefore, the given statement is false because $P(E_1\cap E_2)\neq P(E_1)\cdot P(E_2)$.