A sample space consists of five simple events, E1,2,E3,E4,E5
-I figured out the first part of the question but the second part is where i'm stuck
Part two:
if P(E1)=3P(E2)=0.3, find the probability of the remaining simple events if you know that the remaining simple events are equally likely.

Question

A sample space consists of five simple events, E1,2,E3,E4,E5
-I figured out the first part of the question but the second part is where i'm stuck
Part two:
if P(E1)=3P(E2)=0.3, find the probability of the remaining simple events if you know that the remaining simple events are equally likely.

kirbykirby · Answer

$P(E_1)=0.3$
$2P(E_2)=0.3 \implies P(E_2)=\dfrac{0.3}{2}=0.15$

Now, the sum of all simple events of the sample space must sum to 1 to be a probability distribution. Hence, $P(E_1)+P(E_2)+P(E_3)+P(E_4)+P(E_5)=1$

You already know from the question that $P(E_3)=P(E_4)=P(E_5)$ since they are equally likely... you could essentially say that is the same as $3P(E_3)$. Substitute that into the previous equation :)

kirbykirby · Answer

Ok I misread, it actually says $3P(E_2)=0.3 \implies P(E_2)=\dfrac{0.3}{3}=0.1$