# Simple Probability decoding a message (need clarification about method

• royalewithchz
That makes sense. My mistake was not considering the possibility that all three may not decode the message.
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?

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))

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 $\cup$ 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 $\cap$ 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 $\cap$ 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 $\cup$ B $\cup$ C). This can be calculated in more than one way.

One way to calculate P(A $\cup$ B $\cup$ 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 $\cup$ B $\cup$ C) = P(A $\cap$ B' $\cap$ C') + P(A' $\cap$ B $\cap$ C') + P(A' $\cap$ B' $\cap$ C) + P(A $\cap$ B $\cap$ C') + P(A $\cap$ B' $\cap$ C) + P(A' $\cap$ B $\cap$ C) + P(A $\cap$ B $\cap$ 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 $\cup$ B $\cup$ 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 $\cup$ B $\cup$ C) = 1 - P((A $\cup$ B $\cup$ 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' $\cap$ B' $\cap$ C'). Thus we have:
P(A $\cup$ B $\cup$ C) = 1 - P(A' $\cap$ B' $\cap$ C') = 1 - 5/6*7/8*5/7 = 1 - 175/336 = 161/36 = 23/48.

Thanks Tycon, I really understood your explanation.

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

royalewithchz said:

## 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?

$$\text{P(not decoded)} = (1-p_a)(1-p_b)(1-p_c)$$
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.

Thanks!

Last edited:

## 1. What is Simple Probability decoding and how does it work?

Simple Probability decoding is a method used to decipher a message that has been encoded using a simple substitution cipher. It involves analyzing the frequency of letters or symbols in the encoded message and comparing it to the expected frequency based on the language being used. By identifying the most frequently occurring symbols and their corresponding letters or words, the message can be decoded.

## 2. How accurate is Simple Probability decoding in deciphering a message?

The accuracy of Simple Probability decoding depends on the length and complexity of the message, as well as the language being used. In general, the longer the message and the more frequently occurring symbols there are, the more accurate the decoding will be. However, it is not foolproof and may not work for messages with very short or unique symbols.

## 3. Can Simple Probability decoding be used for any language?

Yes, Simple Probability decoding can be used for any language as long as there is a known expected frequency of letters or symbols for that language. However, it may be more challenging for languages with a larger alphabet or those that use non-Latin characters.

## 4. Are there any limitations to using Simple Probability decoding?

One limitation of Simple Probability decoding is that it relies on the assumption that the frequency of letters or symbols in the encoded message is similar to the expected frequency in the language being used. If the message has been encoded using a different method or if the language is not known, the decoding may not be accurate.

## 5. Are there any other methods that can be used to decode a message?

Yes, there are many other methods for decoding a message, such as frequency analysis, brute force, and computer algorithms. Each method has its own advantages and limitations, and the most effective method will depend on the specific message and circumstances.

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