Respuesta :
I am not sure about the probability but I cant tell you these events are independent because u don't need to roll a five for a coin to land on heads
Let:
F - 5 on a six-sided die,
H - coin lands on heads
We also know that:
[tex]P(F)=\dfrac{1}{6}\qquad\qquad P(H)=\dfrac{1}{2}\qquad\qquad P(F\cap H)=\dfrac{1}{12} [/tex]
Calculate:
[tex]P(H|F)=\dfrac{P(F\cap H)}{P(F)}=\dfrac{\frac{1}{12}}{\frac{1}{6}}=\dfrac{1\cdot6}{12\cdot1}=\dfrac{6}{12}=\boxed{\dfrac{1}{2}}[/tex]
In this case events are independent when:
[tex]P(H|F)=P(H)[/tex]
Both probabilities equals [tex]\dfrac{1}{2}[/tex], so H and F are independent.
F - 5 on a six-sided die,
H - coin lands on heads
We also know that:
[tex]P(F)=\dfrac{1}{6}\qquad\qquad P(H)=\dfrac{1}{2}\qquad\qquad P(F\cap H)=\dfrac{1}{12} [/tex]
Calculate:
[tex]P(H|F)=\dfrac{P(F\cap H)}{P(F)}=\dfrac{\frac{1}{12}}{\frac{1}{6}}=\dfrac{1\cdot6}{12\cdot1}=\dfrac{6}{12}=\boxed{\dfrac{1}{2}}[/tex]
In this case events are independent when:
[tex]P(H|F)=P(H)[/tex]
Both probabilities equals [tex]\dfrac{1}{2}[/tex], so H and F are independent.