Simple Probability decoding a message (need clarification about method

[royalewithchz]

Homework Statement

Three people work independently at deciphering a message sent in code. the respective probabilities are

persion a: 1/6
person b: 1/8
person c: 2/7

what is the probability that the message will be decoded

Homework Equations

(A or B ) = Pr(A) + Pr(b) - Pr(A and B)

A' = 1 - A

The Attempt at a Solution

My first solution was:
decoded message = Pr(A) + Pr(b) + Pr(C) - ( Pr(A)Pr(b)Pr(c) + Pr(A)Pr(B) + Pr(A)Pr(c) + Pr(B)Pr(c))

and I thought it made sense, apparently the answer is not the same. The actual answer is taking the inverse

decoded = ( 1 - Pr(A) ) + ( 1 - Pr(B) ) + ( 1 - Pr(C) )

I understand this method, But I was curious why doesn't my method work?

 

[royalewithchz]

solved.

should be:

Pr(A) + Pr(b) + Pr(C) + Pr(A)Pr(b)Pr(c) - (Pr(A)Pr(B) + Pr(A)Pr(c) + Pr(B)Pr(c))

 

[tycoon515]

Probability

First, let's review basic probability rules.
If we have two discrete random variables that represent random phenomena A and B, then:
P(A [itex]\cup[/itex] B) = P(A) + P(B)
This says that the probability of either A or B occurring (read "A union B") is the sum of their respective probabilities. This is also the probability that at least one of them occurs.

What you had written is a little different. P(A) + P(B) - P(A [itex]\cap[/itex] B) is used to find the probability that A or B occurs, but not both. This is also the probability that exactly one of them occurs.

P(A [itex]\cap[/itex] B) = P(A)*P(B)
This says that the probability of A and B occurring (read "A intersect B") is the product of their respective probabilities.

P(A') = 1 - P(A)
This says that the probability that an event does not occur (its complement) is equal to one minus the probability that the event does occur.

Three people work independently at decoding a message. They will have decoded it as long as at least one of them decodes it. We are interested in the probability that at least one of them decodes it, represented by P(A [itex]\cup[/itex] B [itex]\cup[/itex] C). This can be calculated in more than one way.

One way to calculate P(A [itex]\cup[/itex] B [itex]\cup[/itex] C) is to consider the outcomes where at least one of them decodes it:
A only; B only; C only; A and B; A and C; B and C; A, B, and C
Sum the probabilities of these events:
P(A [itex]\cup[/itex] B [itex]\cup[/itex] C) = P(A [itex]\cap[/itex] B^' [itex]\cap[/itex] C') + P(A' [itex]\cap[/itex] B [itex]\cap[/itex] C') + P(A' [itex]\cap[/itex] B' [itex]\cap[/itex] C) + P(A [itex]\cap[/itex] B [itex]\cap[/itex] C') + P(A [itex]\cap[/itex] B' [itex]\cap[/itex] C) + P(A' [itex]\cap[/itex] B [itex]\cap[/itex] C) + P(A [itex]\cap[/itex] B [itex]\cap[/itex] C)
Given P(A) = 1/6, P(B) = 1/8, and P(C) = 2/7, it follows that P(A') = 5/6, P(B') = 7/8, and P(C') = 5/7.
Applying the rule for the intersection of events, this becomes:
P(A [itex]\cup[/itex] B [itex]\cup[/itex] C) = 1/6*7/8*5/7 + 5/6*1/8*5/7 + 5/6*7/8*2/7 + 1/6*1/8*5/7 + 1/6*7/8*2/7 + 5/6*1/8*2/7 + 1/6*1/8*2/7 = 35/336 + 25/336 + 70/336 + 5/336 + 14/336 + 10/336 + 2/336 = 161/336 = 23/48

Another way to find your probability is to recognize that the probability that A or B or C occurs equals one minus the probability that not ( A or B or C ) occurs.
P(A [itex]\cup[/itex] B [itex]\cup[/itex] C) = 1 - P((A [itex]\cup[/itex] B [itex]\cup[/itex] C)')
The probability that not (A or B or C) occurs equals the probability that A does not occur, B does not occur, and C does not occur, or P(A' [itex]\cap[/itex] B' [itex]\cap[/itex] C'). Thus we have:
P(A [itex]\cup[/itex] B [itex]\cup[/itex] C) = 1 - P(A' [itex]\cap[/itex] B' [itex]\cap[/itex] C') = 1 - 5/6*7/8*5/7 = 1 - 175/336 = 161/36 = 23/48.

 

[royalewithchz]

Thanks Tycon, I really understood your explanation.

Now i know how to do it 3 different ways, really neat!

 

[Ray Vickson]

Homework Statement

Three people work independently at deciphering a message sent in code. the respective probabilities are

persion a: 1/6
person b: 1/8
person c: 2/7

what is the probability that the message will be decoded

Homework Equations

(A or B ) = Pr(A) + Pr(b) - Pr(A and B)

A' = 1 - A

The Attempt at a Solution

My first solution was:
decoded message = Pr(A) + Pr(b) + Pr(C) - ( Pr(A)Pr(b)Pr(c) + Pr(A)Pr(B) + Pr(A)Pr(c) + Pr(B)Pr(c))

and I thought it made sense, apparently the answer is not the same. The actual answer is taking the inverse

decoded = ( 1 - Pr(A) ) + ( 1 - Pr(B) ) + ( 1 - Pr(C) )

I understand this method, But I was curious why doesn't my method work?

[tex] \text{P(not decoded)} = (1-p_a)(1-p_b)(1-p_c) [/tex]
because in order to not be decoded, all three must fail at decoding, and these are independent. This gives the same answer that you get after correcting your expression.

 

[royalewithchz]

Thanks!