Conditional Probabilty

**Revilo** · October 27th 2007, 06:10 AM

First little piggy is 25% likely to go to market if second little piggy goes. Second little piggy is 50% likely to go to if first little piggy goes. Sadly at least one of them must go to market. How likely are they both to go?

So far i have:
Let A = First little piggy goes to market
B = Second little piggy goes to market

P(A given B) = 0.25
P(B given A) = 0.5
P(A union B) = P(A) + P(B) - P(A intersection B)

Is the P(A union B) = 1 as atleast one must go, so this is always true?
Does anyone have any ideas, i'm really stuck.

**kalagota** · October 28th 2007, 01:11 AM

Originally Posted by Revilo

First little piggy is 25% likely to go to market if second little piggy goes. Second little piggy is 50% likely to go to if first little piggy goes. Sadly at least one of them must go to market. How likely are they both to go?

So far i have:
Let A = First little piggy goes to market
B = Second little piggy goes to market

P(A given B) = 0.25
P(B given A) = 0.5
P(A union B) = P(A) + P(B) - P(A intersection B)

Is the P(A union B) = 1 as atleast one must go, so this is always true?
Does anyone have any ideas, i'm really stuck.

yeah, it is correct..
but, i think, you should not look for the union, instead for the intersection..
note that P(A|B)= P(A int B) / P(B) [also, P(B|A) = P(A int B) / P(A) ]; this theorem should help you find the answer.

**Soroban** · October 28th 2007, 08:51 PM

Hello, Revilo!

First little piggy is 25% likely to go to market if second little piggy goes.
Second little piggy is 50% likely to go to if first little piggy goes.
Sadly at least one of them must go to market.
How likely are they both to go?

All your ideas are good ones . . .

Bayes' Theorem says: . $P(A|B) \:=\:\frac{P(A \wedge B)}{P(B)}$

We are given:

. . $P(1^{st}|2^{nd}) \:=\:\frac{1}{4}\quad\Rightarrow\quad \frac{P(1^{st} \wedge 2^{nd})}{P(2^{nd})}\: = \: \frac{1}{4}\quad\Rightarrow\quad P(1^{st} \wedge 2^{nd}) \:=\:\frac{1}{4}\!\cdot\!P(2^{nd})$ .[1]

. . $P(2^{nd}|1^{st}) \:=\:\frac{1}{2}\quad\Rightarrow\quad \frac{P(2^{nd} \wedge 1^{st})}{P(1^{st})} \:=\:\frac{1}{2}\quad\Rightarrow\quad P(1^{st} \wedge 2^{nd}) \:=\:\frac{1}{2}\!\cdot\!P(1^{st})$ . [2]

Equate [1] and [2]: . $\frac{1}{4}\cdot P(2^{nd}) \:=\:\frac{1}{2}\cdot P(1^{st}) \quad\Rightarrow\quad P(2^{nd}) \:=\:2\cdot P(1^{st})$ . [3]

Since at least one of them goes to the market: . $P(1^{st} \vee 2^{nd}) \:=\:1$

We have: . $P(1^{st} \vee 2^{nd}) \;=\;P(1^{st}) + P(2^{nd}) - P(1^{st} \wedge 2^{nd}) \;=\;1$ . [4]

We already know that:
. . [3] $\; P(2^{nd}) \: = \: 2\cdot P(1^{st})$
. . [2] $\; P(1^{st} \wedge 2^{nd}) \: = \: \frac{1}{2}\cdot P(1^{st})<br />$

Substitute into [4]: . $P(1^{st}) + 2\cdot P(1^{st}) -\frac{1}{2}\cdot P(1^{st}) \;=\;1\quad\Rightarrow\quad \frac{5}{2}\cdot P(1^{st}) \:=\:1$

Therefore: . $P(1^{st}) \:= \:\frac{2}{5}\qquad P(2^{nd}) \: = \: \frac{4}{5} \qquad \boxed{P(1^{st} \wedge 2^{nd}) \: = \: \frac{1}{5}}$

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