# radio locators: prob. of each locator finding object is 0.84, 0.90, 0.80, 0.95

#### Plamen

##### New member
4 radio locators with different constructions. probability of finding and object with the help of the first is 0.84, second – 0.90, third – 0.80 and forth – 0.95.

A viewer turns on one radio locator. What is the probability of finding an object ?

I'm guessing that this is the right answer(all of the probabilities added and then divided by 4) -> (0.84 + 0.90 + 0.80 + 0.95) / 4 = 3.49 / 4 = 0.87 ? Is that correct ?

But then I am asked the question - After turning on one of the radio locators an object is registered. what is the probability of this being the second radio locator ?

I know the formula for this is P(A|B) = P(AB) / P(B). In this case B should be the condition being met - the radio locator found an object. I don't know how to multiply A and B ?

Any idea or guidance ?

#### Dr.Peterson

##### Elite Member
4 radio locators with different constructions. probability of finding and object with the help of the first is 0.84, second – 0.90, third – 0.80 and forth – 0.95.

A viewer turns on one radio locator. What is the probability of finding an object ?

I'm guessing that this is the right answer(all of the probabilities added and then divided by 4) -> (0.84 + 0.90 + 0.80 + 0.95) / 4 = 3.49 / 4 = 0.87 ? Is that correct ?
If the viewer randomly turned exactly one on, then you are right. Can you justify your answer by giving a reason, rather than just guess? (The fact that you are right does not imply that your thinking was correct.)

But then I am asked the question - After turning on one of the radio locators an object is registered. what is the probability of this being the second radio locator ?

I know the formula for this is P(A|B) = P(AB) / P(B). In this case B should be the condition being met - the radio locator found an object. I don't know how to multiply A and B ?
In your notation, P(AB) means "probability of A and B". It is not multiplication.

What are your events A and B?

#### Plamen

##### New member
Why i think this way

I think that we have to take all four devices and divide them by 4 because they all have equal chances of getting picked. Exactly one is chosen at random.

A is the probability of the second device being selected. B is that the second device detected something.
So here I'm thinking i have to multiply the numbers (0.87 * P(B)) / P(B). which i guess it should multiply P(B) and then divide it ? So the answer should just stay 0.87 ?
I just don't know what to make of this P(B) when this device has a chance of detecting something and then it happens. How is that affecting the formula ?

#### Dr.Peterson

##### Elite Member
I think that we have to take all four devices and divide them by 4 because they all have equal chances of getting picked. Exactly one is chosen at random.
Is there a theorem or formula you have learned that applies to this? "I think" can be very dangerous.

A is the probability of the second device being selected. B is that the second device detected something.
So here I'm thinking i have to multiply the numbers (0.87 * P(B)) / P(B). which i guess it should multiply P(B) and then divide it ? So the answer should just stay 0.87 ?
I just don't know what to make of this P(B) when this device has a chance of detecting something and then it happens. How is that affecting the formula ?
Why did you multiply? Is it because you associate "and" with multiplying? That is true only when two events are independent. That is, P(A and B) = P(A)*P(B) only when A and B are independent events, which is not true here.

What you want is "the probability that the second locator detected the object, given that an object is detected after turning on one of the locators". So in P(A | B), A is "the second locator detected the object", and B is "an object is detected", all on the understanding that only one locator is on. Event B doesn't "know" which locator it was; your definition was incorrect.

So P(A and B) is the probability that the second locator detected an object (that is, an object was detected, and it was the second locator that did it), and P(B) is the probability that an object was detected (by any locator) - which is what the first part of the problem asked.

#### Plamen

##### New member
Why i think this way

Well For the first one there is no formula I'm using just my common sense.
Is there another way to think about this problem or maybe there is a formula I'm missing here?

I multiplied because i know that or is logical +, and is logical multiplication. But guess that is not the case here.
So if an object was detected, and it was the second locator that did it is A and B.
By your explanations i should just have a solution with P(A|B) = 0.90 / 0.87 ?

#### Dr.Peterson

##### Elite Member
Well For the first one there is no formula I'm using just my common sense.
Is there another way to think about this problem or maybe there is a formula I'm missing here?

I multiplied because i know that or is logical +, and is logical multiplication. But guess that is not the case here.
So if an object was detected, and it was the second locator that did it is A and B.
By your explanations i should just have a solution with P(A|B) = 0.90 / 0.87 ?
No, this would be an impossible probability, since it is greater than 1.

I guess what I'm really hoping for from you is an indication of what you have learned about probability. There is a formula for the first question: P(A) = P(A|B1)*P(B1) + ... + P(A|Bn)*P(Bn) , where B1 through Bn are mutually exclusive events that together cover the entire sample space (a partition). In your case, they are "first locator is chosen, second locator is chosen, ...", and the probability of each of them is 1/4; so the formula amounts to [P(A|B1) + ... + P(A|Bn)]*(1/4), which is your average. Have you learned anything of the sort, that you would be expected to use for this problem?

"Or" being addition and "and" being multiplication is a general statement that has to be made precise; Each has a general form (more complex) and a special case (mutually exclusive and independent, respectively) in which it is as simple as it can be. Surely you have learned these fuller forms. This is a good place to see what I am referring to.

Here is one site that includes these and many other rules, including the "law of total probability", listed under Bayes' Rule, which is what I mentioned for the first part. The general cases are stated succinctly here.