# Conditional Probability Problem

• Aug 30th 2008, 03:13 AM
ban26ana
Conditional Probability Problem
Problem: Balls are drawn (one at a time) from an urn containing M balls numbered from 1 to M. The ball is kept if it is numbered 1, and returned to the urn otherwise. What is the probability that the second ball drawn is numbered 2?

Here is what I have figured. If the ball numbered 1 is chosen first:
P(2) = 1/M * 1/(M-1)

If the ball numbered 1 is not chosen first:
P(2) = (M-1)/M * 1/M

My question is: Are these added together to get the overall probability? Meaning, is the final answer:
[1/M * 1/(M-1)] + [(M-1)/M * 1/M] ? Thanks!
• Aug 30th 2008, 04:04 AM
Moo
Hello !
Quote:

Originally Posted by ban26ana
Problem: Balls are drawn (one at a time) from an urn containing M balls numbered from 1 to M. The ball is kept if it is numbered 1, and returned to the urn otherwise. What is the probability that the second ball drawn is numbered 2?

Here is what I have figured. If the ball numbered 1 is chosen first:
P(2) = 1/M * 1/(M-1)

If the ball numbered 1 is not chosen first:
P(2) = (M-1)/M * 1/M

My question is: Are these added together to get the overall probability? Meaning, is the final answer:
[1/M * 1/(M-1)] + [(M-1)/M * 1/M] ? Thanks!

Yup (Happy)

Do it with letters if you want to prove it =)
Let A be the event "The first ball drawn is numbered 1". $\displaystyle \bar{A}$ is the complementary event "the first ball drawn is not numbered 1".
Let B be the event "The 2nd ball drawn is numbered 2".

You're looking for $\displaystyle P(B)$

There's a formula that would let you write :

$\displaystyle P(B)=P(B/A)P(A)+P(B/\bar{A})P(\bar{A})$ (*)
Which gives exactly what you've written.

Note, $\displaystyle P(B/A)$ represents "the probability of B to happen if A happened before".

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(*) This can be proved "easily" (maybe it can interest Chris as well (Rofl)) :
$\displaystyle B=(B \cap A) \cup (B \cap \bar{A})$

We know that $\displaystyle P(M \cup N)=P(M)+P(N)-P(M \cap N)$

Thus $\displaystyle P(B)=P(B \cap A)+P(B \cap \bar{A})-P((B \cap A) \cap (B \cap \bar{A}))$

~~~~~~~~~~~~~~~~~~
$\displaystyle (B \cap A) \cap (B \cap \bar{A})=\underbrace{(B \cap B)}_{B} \cap (A \cap \bar{A})$ (property of commutativity and associativity)
But $\displaystyle A \cap \bar{A}=\emptyset$ because they're complementary !
Thus $\displaystyle (B \cap A) \cap (B \cap \bar{A})=B \cap \emptyset =\emptyset$
~~~~~~~~~~~~~~~~~~

$\displaystyle \implies P(B)=P(B \cap A)+P(B \cap \bar{A})-\underbrace{P(\emptyset)}_{=0}=P(B \cap A)+P(B \cap \bar{A})$

By the formula of conditional probability, we have : $\displaystyle P(B \cap A)=P(B/A)P(A)$ and $\displaystyle P(B \cap \bar{A})=P(B/\bar{A})P(\bar{A})$

---> $\displaystyle \boxed{P(B)=P(B/A)P(A)+P(B/\bar{A})P(\bar{A})}$

Uuuuuh clear enough ?
Long ? Clearly enough :D
• Aug 30th 2008, 10:35 AM
ban26ana
Thank you so much. Yup, confusing enough. LOL
• Aug 30th 2008, 08:44 PM
mr fantastic
Quote:

Originally Posted by ban26ana
Problem: Balls are drawn (one at a time) from an urn containing M balls numbered from 1 to M. The ball is kept if it is numbered 1, and returned to the urn otherwise. What is the probability that the second ball drawn is numbered 2?

Here is what I have figured. If the ball numbered 1 is chosen first:
P(2) = 1/M * 1/(M-1)

If the ball numbered 1 is not chosen first:
P(2) = (M-1)/M * 1/M

My question is: Are these added together to get the overall probability? Meaning, is the final answer:
[1/M * 1/(M-1)] + [(M-1)/M * 1/M] ? Thanks!

A tree diagram gives the answer quickly, easily and with little confusion ......

The first two branches are the events "1" and "Not 1". The second two branches (that branch out from each of the first two branches) are the events "2" and "Not 2".