probability in multiple events when tossing a single die

[letsgetaway]

I have a probability that overlaps several events. I'm not so sure how to calculate the probability of the simple event because multiple events satisfy yet don't satisfy the event. I hope I'm not confusing anyone yet. Here's the problem...

events involving a toss of a single die
all simple events have a 1/6 probability

a: observe a 2 ... a = {2}
b: observe an even number ... b = {2, 4, 6)
c: observe a number greater than 2 ... c = {3, 4, 5, 6}
d: observe both a and b ... d = {2}
e: observe a or b or both ... e = {2, 4, 6}
f: observe both a and c ... f = {} "not possible"

I need help finding P(C).
P(C) the book says the answer is 2/3. I don't know how they got that answer.
I thought P(C) = a portion of Event B excluding 2 + a portion of Event E excluding 2 = 1/3

 

[wjm11]

Re: probability in multiple events

events involving a toss of a single die
all simple events have a 1/6 probability

a: observe a 2 ... a = {2}
b: observe an even number ... b = {2, 4, 6)
c: observe a number greater than 2 ... c = {3, 4, 5, 6}
d: observe both a and b ... d = {2}
e: observe a or b or both ... e = {2, 4, 6}
f: observe both a and c ... f = {} "not possible"

I need help finding P(C).
P(C) the book says the answer is 2/3. I don't know how they got that answer.

P(greater than 2) = P(3) + P(4) + P(5) + P(6) = 1/6 +1/6 +1/6 + 1/6 = 4/6 = 2/3.

Alternatively, P(greater than 2) = 1 – P(not greater than 2) = 1 – [P(1) + P(2)] = 1 – [1/6 + 1/6] = 1 – 2/6 = 2/3.

Hope that helps.

 

[Loren]

Your abbreviations leave something to be desired. You refer to P(C). No reference to C beforehand. Possibly, you mean P(c) which, I assume, means "what is the probability of event described in line c occurring?". If that is what is meant, you have a sample space of 6. That is you have six different possible outcomes, namely, 1, 2, 3, 4, 5, or 6. Line c describes 4 possible outcomes --- 3, 4, 5, or 6. The probability of line c described outcome divided by the total possible outcomes is 4/6. I think that reduces to the desired answer.

 

[letsgetaway]

Your abbreviations leave something to be desired. You refer to P(C). No reference to C beforehand. Possibly, you mean P(c) which, I assume, means "what is the probability of event described in line c occurring?". If that is what is meant, you have a sample space of 6. That is you have six different possible outcomes, namely, 1, 2, 3, 4, 5, or 6. Line c describes 4 possible outcomes --- 3, 4, 5, or 6. The probability of line c described outcome divided by the total possible outcomes is 4/6. I think that reduces to the desired answer.

line C was the same as P(C). I used c and C interchangeably. I didn't think that made a difference.

Thanks for clarifying how the answer was obtained. I do see my error now. I was looking at all my events each with a total of "1/6"