Having been reprimanded for posting a question on the wrong site, I hope I'm not transgressing this

Xiomara Poole

Xiomara Poole

Answered question

2022-05-30

Having been reprimanded for posting a question on the wrong site, I hope I'm not transgressing this time. In the addition rule for non mutually exclusive events P ( A B ) = P ( A ) + P ( B ) P ( A B ), while P ( A B ) can be determined by observation, why cannot it always be calculated from P ( A ) P ( B )?

Example: A lottery box contains 50 lottery tickets numbered 1 to 50. If a lottery ticket is drawn at random, what is the probability that the number drawn is a multiple of 3 or 5? P ( X Y ) = P ( X ) + P ( Y ) P ( X Y ) Therefore,
P ( X U Y ) = 8 / 25 + 1 / 5 3 / 50 = ( 16 + 10 3 ) / 50
= 23 / 50
But using P ( X Y ) = P ( X ) P ( Y ) = 8 / 25 1 / 5 = 8 / ( 25 5 ) = 8 / 125 which is NOT 3/50?
There are situations where P ( X Y ) = P ( X ) x P ( Y ) works perfectly, but not in others.

Answer & Explanation

Louis Lawrence

Louis Lawrence

Beginner2022-05-31Added 10 answers

If X and Y are dependent then P ( X Y ) = P ( X ) P ( Y | X )
P ( X ) = 8 25
The favorable outcomes of X are 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , 48
The number of outcomes of X is 16.
Three of them are divisible by 5. Thus P ( Y | X ) = 3 16
Therefore P ( X Y ) = 8 25 3 16 = 3 50

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school probability

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?