Probability

Question

Consider a sample space S=(e₁, e₂, e_3, e₄) with P(e₁)=2P(e₃), P(e₂,e₃)=3P(e₁) and P(e₄)=3P(e₃). Find P(e₁), P(e₂), P(e₃), and P(e₄).

 

Answer 1

I'll assume that P(e₂, e₃) = 3P(e₁) means P(e₂) = 3P(e₁) and P(e₃) = 3P(e₁).

 

 

 

All together, the probabilities of the events in a sample space must add up to 1. So:

 

P(e₁) + P(e₂) + P(e₃) + P(e₄) = 1

 

 

 

Let's convert everything into P(e₁) and solve for that:

 

P(e₁) + 3P(e₁) + 3P(e₁) + 3P(e₃) = 1

P(e₁) + 3P(e₁) + 3P(e₁) + 3(3P(e₁)) = 1

7P(e₁) + 9P(e₁) = 1

16P(e₁) = 1

P(e₁) = 1/16

 

That means, plugging back in, that:

 

P(e₂) = P(e₃) = 3P(e₁) = 3(1/16) = 3/16

P(e₄) = 3P(e₃) = 3(3/16) = 9/16

 

 

 

Let's check our work by adding the probabilities together:

 

1/16 + 3/16 + 3/16 + 9/16 = 16/16 = 1

 

That checks out!