Thread: Mutuallly exclusive events

John and his friend plan to travel to North Dakota during winter break. The probability that they go by car is 2/3 and the probability that they go by plane is 1/5. What is the probability that they go to Florida by car or plane?

These are independent events so P(A)+P(B)

2/3 + 1/5 = 13/15??

Originally Posted by terminator

[FONT=Arial] John and his friend plan to travel to North Dakota during winter break. The probability that they go by car is 2/3 and the probability that they go by plane is 1/5. What is the probability that they go to Florida by car or plane?

The answer is zero. They plan to go to North Dakota not to Florida.

Originally Posted by terminator

John and his friend plan to travel to North Dakota during winter break. The probability that they go by car is 2/3 and the probability that they go by plane is 1/5. What is the probability that they go to Florida by car or plane?

These are independent events so P(A)+P(B) what????????

2/3 + 1/5 = 13/15??

NO matter where John and his friend go, you first need to know that Independent events and mutually exclusive events are NOT the same!

I know I make mistakes, but I've learned a lot from you guys. Thanks for all the feedback provided promptly.

These should be mutually exclusive events.
P(A) + P(B) = 13/15

Originally Posted by harish21

NO matter where John and his friend go, you first need to know that Independent events and mutually exclusive events are NOT the same!

@harish21, What are we to do? Are we expected to be mind readers?
Unless the poster is clear about the meaning of the post, chaos will follow.
Why do you think that a good guess is better than asking for clarification?

Originally Posted by Plato

@harish21, What are we to do? Are we expected to be mind readers?
Unless the poster is clear about the meaning of the post, chaos will follow.
Why do you think that a good guess is better than asking for clarification?

I think harish21 was addressing this misconeption revealed by the OP:

These are independent events so P(A)+P(B)

This is a seperate issue to the actual question that was posted but is still relevant to the original post.

Edit: Thread closed: See here: http://www.mathhelpforum.com/math-he...tml#post569935

(By the way, post #2 in this thread is my favourite post of the year ....)