A few people gather together to play a game in which one needs to roll 3 six-sided dice. One person notices that the sum of the number of spots on three dice comes up as 11 (the event E1) more often than it does 12 (the event E2) even though it looks like it should be even. This person argues as follows: E1 occurs in just six ways: {(1,4,6),(2,3,6),(1,5,5),(2,4,5),(3 ,3,5),(3,4,4)}, and E2 occurs also in just six ways: {(1,5,6),(2,4,6),(3,3,6),(2,5,5),(4 ,4,4),(3,4,5)}.

Therefore E1 and E2 have the same probability P(E1) = P(E2). Why isn’t it?

Yes, P(E1)=P(E2)

Step-by-step explanation:

We are given that

Number of dice=3

Total outcomes of 1 die=6

Therefore,

Total number of outcomes =

E1={{(1,4,6),(2,3,6),(1,5,5),(2,4,5),(3,3,5),(3,4,4)}

E2={(1,5,6),(2,4,6),(3,3,6),(2,5,5),(4,4,4),(3,4,5)}

We have to show that E1 and E2 have the same probability P(E1) = P(E2).

Probability,

Using the formula

Hence, P(E1)=P(E2)

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.

she was supposed to add the hours(38.5) and multiply the amount of money she earned(7.50) resulting in $288.75 .

correct me if i'm wrong and if it is i'm truly sorry.