Two events ** E **and

*are*

**F****independent**if the occurrence of

**in a probability experiment does not affect or alter the probability of event**

*E***occuring. In other words, knowing that**

*F**occurred does not give any additional information about whether*

**E****will or will not occur; knowing that**

*F***occurred does not give any additional information about the occurance of**

*F***. Therefore, events**

*E***and**

*E***are independent if they are totally unrelated. For example, if you are flipping a fair coin (this means the probability of getting a heads is 50% and getting a tails is 50%), does knowing that you just flipped a tails tell us anything about what will happen the next time we flip the coin? No! The coin has no “memory” to speak of. Even if you flipped 10 heads in a row, the probability of flipping heads on the 11**

*F*^{th}toss is still 50%.

If the two events are not independent, then they are said to be **dependent**. If two events are dependent, it does not mean that they completely rely on each other; it just means that they are not independent of each other. In other words, there is some kind of relationship between ** E** and

*, even if it is just a very small relationship. For example, you are asked to pull one card from a standard deck of 52 cards. Let*

**F****and**

*E*= {red card}**. Suppose you pull a red card from the deck. Does knowing this provide any information about whether**

*F*= {black card}*occurred? Yes! If we pulled a red card, then we know we didn’t pull a black card, so therefore*

**F****could not have occurred!**

*F*Let’s run a different experiment by pulling two cards from a standard deck without replacement. If the first card pulled is a red card, does that change the probability that we will pull a black card for the second card? Most definitely, because now there is one fewer red card in the deck, which actually increases the probability that the second card is black (even though the change in probability is small).

#### The Multiplication Rule for Independent Events

The Multiplication Rule for independent events states:

*P* (*E* and *F*) = *P* (*E*) ⋅ *P* (*F*)

Thus we can find ** P (E and F)** if we know

**and**

*P*(*E*)**. This is also true for more than two independent events. So if**

*P*(*F*)**are all independent from each other, then:**

*E, F, G, …**P* (*E* and *F* and *G* and ⋯) = *P *(*E*) ⋅ *P* (*F*) ⋅ *P *(*G*) ⋯

The ELISA is a test to determine whether the HIV antibody is present in a patient’s blood. The test is 99.5% effective. This means that the test will accurately come back negative if the HIV antibody is not present. The probability of a test coming back positive when the antibody is not present (known as a *false positive*) is **100% – 99.5% = 0.5% = 0.005**. Suppose the ELISA is given to 5 randomly selected people who do not have the HIV antibody.

(a)

What is the probability that the ELISA comes back negative for all five people? First, testing each individual with the ELISA is an independent event, because knowing the results of the test for one person gives us no information about what the result will be for the next person. Therefore:

*P* (all 5 tests are negative) = (0.995) ⋅ (0.995) ⋅ (0.995) ⋅ (0.995) ⋅ (0.995)

= (0.995)^{5}

≈ 0.9752

Therefore, there is a 97.52% that all 5 individuals will test negative for the HIV antibody when all 5 patients are indeed HIV-negative.

(b)

What is the probability that the ELISA comes back positive for *at least* one of the five people? First of all, “at least one” means 1 or 2 or 3 or 4 or 5 of the people receives a positive test. Another way to say “all 5 tests are negative” is “none of the 5 tests is positive.” In symbols, if ** E = {all 5 have a negative ELISA}**, then we could also just as well say

**. Therefore, we know the compliment of**

*E*= {none of the 5 have a positive ELISA}*to be*

**E****=**

*E*^{c }**{at least one of the 5 has a positive ELISA}**. Using the fact that

**, we see:**

*P*(*E*) = 1 –^{c}*P*(*E*)*P *(at least one of the 5 tests positive) = 1 – *P *(all 5 have negative tests)

= 1 – (0.995)^{5}

≈ 1 – 0.9752

≈ 0.0248

There is a 2.48% chance of at least one of the 5 individuals getting a false positive reading. This is an usual event (since the probability value is very low), as it should be, as false positive results can cause an individual undue emotional stress and result in additional (often extremely expensive) testing.