Question

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

Answer 1

A=2,4,6,B=1,2,3

A∩B=2

P(A)=36=12,P(B)=36=12

P(A∩B)=16

Hence P(A∩B)≠P(A)P(B)

Hence A and B are not independent event.

Answer 2

No, A and B are not independent.

Step-by-step explanation:

Given the dice marked 1,2,3 in red and 4,5,6 in green

sample space is  $\left | \textrm{s} \right |=6$

 Let A be the event of getting the number even

  $\textrm{A}=\left \{ 2,4,6 \right \}$

 $\left | \textrm{A} \right |=3$

and B be the event of getting the number is Red

$\textrm{B}=\left \{ 1,2,3 \right \}$

$\left | \textrm{B} \right |=3$

$\textrm{P(A)}=\dfrac{\left | \textrm{A} \right |}{\left | \textrm{S} \right |}=\dfrac{3}{6} =\dfrac{1}{2}$

and $\textrm{P(B)}=\dfrac{\left | \textrm{B} \right |}{\left | \textrm{S} \right |}=\dfrac{3}{6}=\dfrac{1}{2}$

$\textrm{A}\bigcap \textrm{B}=\left \{ 2,4,6 \right \}\bigcap \left \{ 1,2,3 \right \}=\left \{ 2 \right \}$

$\textrm{P}(\textrm{A}\bigcap \textrm{B})=\frac{\left | \textrm{A}\bigcap \textrm{B} \right |}{\left | \textrm{S}\right |}$$=\dfrac{1}{6}$

We have to show it is independent or not

i.e $\textrm{P}(\textrm{A}\bigcap \textrm{B})=\textrm{P}(\textrm{A}). \textrm{P}(\textrm{B})$

$\textrm{P}(\textrm{A}). \textrm{P}(\textrm{B})=\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{4}\neq \textrm{P}(\textrm{A}\bigcap \textrm{B})$

So A and B are not Independent.