One Dice

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Jack tosses one die. What is the probability of the die landing on 5?

Let me see.

There are 6 numbers on a die: 1, 2, 3, 4, 5, 6.
There is only one number 5 on that list.

Let E = landing on 5

So. P(E) = 1/(total numbers listed).

P(E) = 1/6

Yes?
 
When I saw the title "One Dice" I was ready to open this thread and scream and shout! But then I saw that in the body you write "die". Well done!
 
"Jack tosses one die. What is the probability of the die landing on 5?
Let me see.
There are 6 numbers on a die: 1, 2, 3, 4, 5, 6."
Credit where credit is due.
 


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