Lab Rat Independence

Probability Level 1

A scientist is conducting an experiment with two rats, and she noted the incidence of the following events each day.

$A=\text{The 1st rat receives an extra food pellet for the day}$

$B=\text{The 1st rat runs in the exercise wheel that day}$

$C=\text{The 2nd rat runs in the exercise wheel that day}$

Over the course of the experiment, the scientist recorded the following probabilities:

$\begin{array}{lll} P(A)=0.5 & P(B)=0.2 & P(C)=0.1 \\ P(A\cap B)=0.1 & P(A\cap C)=0.05 & P(B\cap C)=0.02 \\ P(A\cap B\cap C)=0.01 \\ \end{array}$

Are the events mutually independent?

Yes No Not enough information

1 solution

Andy Hayes
Jun 7, 2016

Relevant wiki: Probability - Independent events

If the events are mutually independent, then the probability of each intersection of events will equal the product of the probabilities of those events.

$P(A\cap B)=P(A)\times P(B)=0.1$

$P(A\cap C)=P(A)\times P(C)=0.05$

$P(B\cap C)=P(B)\times P(C)=0.02$

$P(A\cap B\cap C)=P(A)\times P(B)\times P(C)=0.01$

These calculations match the probabilities listed. Therefore, the events are independent.

When you say "If the events are mutually independent, then the probability of each intersection of events will equal the product of the probabilities of those events," what do you mean that the probability of each intersection of events will equal to the product?

I'm assuming that the probability of each intersection refers to the probability of 2 of the events happening together. So when you say they all equal to the product of the probability of those events; how do you know they equal?

0.1 + 0.05 + 0.02 does not equal to 0.01, and 0.1 x 0.05 x 0.02 also does not equal to 0.01. Can you explain this concept to me?

Ryan Huynh - 3 years, 4 months ago

Hi, I understand how you are confused on this. Just like you said, "I'm assuming that the probability of each intersection refers to the probability of 2 of the events happening together." you are right. Now go through that again and see how they calculated it and then think about how you would test for the 3 way intersection in the same logic and the pairwise.

P(A) P(B) [0.5 0.2] = 0.1. This is also the same answer for the intersection of A&B that the scientist calculated in the question. P(B) P(C) [0.2 0.1] = 0.02. This is also the same answer for the intersection of B&C that the scientist calculated in the question. So we can see so far that yes, the product of the probabilities is equal to the intersection in the examples. Try now and see if the product of the probabilities of A,B, and C is equal to the intersection probability that the scientist calculated.

"Hint, don't multiply the probabilities of the intersections because you may be double counting here."

Matthew Williams - 3 years, 3 months ago

Wow... I can't believe it was that simple. Thank you for explaining this to me!

Ryan Huynh - 3 years, 3 months ago

The probability of A multiplied by the probability of B is equal to the probability of A & B.

P(A) * P(C) = P(A & B)

P(B) * P(C) = P(B & C)

So the are certainly mutually pairwise independent. If they were pairwise dependent, then the probability of both items in a pair would be different than the probability of one multiplied by the probability of the other.

Now we just do the same computation for all 3. The probability of all 3, if they are independent, is the probability of each, multiplied with each other. If they are dependent, it would be less, because we would have to subtract out some overlapping part of the venn diagram. Since multipliying the probabilities of each event does equal the probability of all 3 events, they must be mutually independent.

buggle snug - 2 years, 5 months ago

@Buggle Snug I believe you meant P(A) * P(B) = P(A & B) on line 2.

Anthony Lee - 3 months, 3 weeks ago

Lab Rat Independence

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