Mutually-exclusive events?

[thornluke]

Homework Statement

When David goes fishing the probability of him catching a fish of type A is 0.45, catching a fish of type B is 0.75 and catching a fish of type C is 0.2.
David catches four fish.
If the event X is David catching two fish of type A and two other fish, the event Y is David catching two fish of type A and two of type B and the event Z is David catching at least one fish of type C, for each of the pairs of X, Y and Z state whether the two events are mutually exclusive, giving a reason.

Homework Equations

The Attempt at a Solution

Event X = P(A) x P(A) x P(A $\cup$ B) x P(A $\cup$ B)
Event Y = P(A) x P(A) x P(B) x P(B)
Event Z = P(C)

 

[Ray Vickson]

Homework Statement

When David goes fishing the probability of him catching a fish of type A is 0.45, catching a fish of type B is 0.75 and catching a fish of type C is 0.2.
David catches four fish.
If the event X is David catching two fish of type A and two other fish, the event Y is David catching two fish of type A and two of type B and the event Z is David catching at least one fish of type C, for each of the pairs of X, Y and Z state whether the two events are mutually exclusive, giving a reason.

Homework Equations

The Attempt at a Solution

Event X = P(A) x P(A) x P(A $\cup$ B) x P(A $\cup$ B)
Event Y = P(A) x P(A) x P(B) x P(B)
Event Z = P(C)

You are writing nonsense: an *event* and that event's *probability* are not the same thing! So, whether or not events X and Y are mutually exclusive is not governed by values of P(A) or P(B) or P(C). In other words, even if we changed P(A), P(B) and P(C) from 0.45, 0.75 and 0.2 to something else (other than 0 or 1) that would in no way affect whether or not X, Y, Z are mutually exclusive.

RGV

 

[HallsofIvy]

I won't go so far as to say what you have written as your answer is "non-sense", it just doesn't have anything to do with the question! This question is not asking for a numerical answer.

The question is, is it possible for David to catch four fish such that the statements, "David caught two fish of type A and two other fish" and "David caught two fish of type A and two fish of type B" can both be true. If that is possible then X and Y are NOT "mutually exclusive". If it is not possible then they ARE "mutually exclusive".

To determine if X and Z are mutually exclusive, think about whether both statements X= "David caught two fish of type A and two other fish" and Z= "David caught at least one fish of type C" can both be true of the same four fish.

To determine if Y and Z are mutually exclusive, think about whether both statements Y= "David caught two fish of type A and two fish of type B" and Z= "David caught at least one fish of type C" can both be true of the same four fish.

 

[Ray Vickson]

I won't go so far as to say what you have written as your answer is "non-sense", it just doesn't have anything to do with the question! This question is not asking for a numerical answer.

The question is, is it possible for David to catch four fish such that the statements, "David caught two fish of type A and two other fish" and "David caught two fish of type A and two fish of type B" can both be true. If that is possible then X and Y are NOT "mutually exclusive". If it is not possible then they ARE "mutually exclusive".

To determine if X and Z are mutually exclusive, think about whether both statements X= "David caught two fish of type A and two other fish" and Z= "David caught at least one fish of type C" can both be true of the same four fish.

To determine if Y and Z are mutually exclusive, think about whether both statements Y= "David caught two fish of type A and two fish of type B" and Z= "David caught at least one fish of type C" can both be true of the same four fish.

Well, if "nonsense" means "makes no sense", then what he wrote clearly satisfies that: he says things like "Event X = P(A) x P(A) x P(A ∪ B) x P(A ∪ B)", so he is equating an event and a probability. To me, that is like equating a fish and a bicycle. Maybe he did not really mean what he wrote. Perhaps he meant to say P(X) = P(A) x P(A) x P(A ∪ B) x P(A ∪ B), etc. That would at least have meaning (but still be wrong).

It is true that I had not added anything more helpful to my first response, but I had been hoping that just making the OP aware of the error would be enough to help get him started down the right track (for example, by encouraging him to read his textbook, review his course notes, or go online and use Google).

RGV