Intersections of Events and Independence

Dependent and Independent Events:

Consider these two groups of pairs of events.

Group 1

A = {airplane engine fails in flight}

B = {airplane reaches its destination safely}

or (when a fair coin is tossed twice) A = {H is on 1st toss}

B = {H on both tosses}.

Group 2

A = {a coin toss shows heads}

B = {a bridge hand has 4 aces}.

or (when a fair coin is tossed twice) A = {H on 1st toss}

B = {H on 2nd toss}

What do the pairs A, B in each group have in common? In group 1 the events are related so that the occurrence of A affects the chances of B occurring. In group 2, whether A occurs or not has no effect on B’s occurrence.

A

B P(AB)=P(A)P(B)

S

Figure 4.6: Independent Events A, B

We call the pairs in group 1 dependent events, and those in group 2 independent events. We formalize this concept in the mathematical definition which follows.

Definition 6 Events A and B are independent if and only if P (AB) = P (A)P (B). If they are not independent, we call the events dependent.

If two events are independent, then the “size” of their intersection as measured by the probability measure is required to be the product of the individual probabilities. This means, of course, that the intersection must be non-empty, and so the events are not mutually exclusive. For example in the Venn diagram depicted in Figure 4.6, P (A) = 0.3, P (B) = 0.4 and P (AB) = 0.12 so in this case the two events are independent.

For another example, suppose we toss a fair coin twice. Let A = {head on 1st toss} and B = {head on 2nd toss}. Clearly A and B are independent since the outcome on each toss is unrelated to other tosses, so P (A) = ¹₂, P (B) = ¹₂, P (AB) = ¹₄ = P (A)P (B).

However, if we roll a die once and let A = {the number is even} and B = {number > 3} the events will be dependent since

P (A) = 1

2, P (B) = 1

2, P (AB) = P (4 or 6 occurs) = 2

6 6= P (A)P (B).

(Rationale: B only happens half the time. If A occurs we know the number is 2, 4, or 6. So B occurs

2

3 of the time when A occurs. The occurrence of A does affect the chances of B occurring so A and B are not independent.)

When there are more than 2 events, the above definition generalizes to:

Definition 7 The events A₁, A₂, · · · , Aⁿare independent if and only if P (Ai1, Ai2, · · · , Aⁱk) = P (Ai1)P (Ai2) · · · P (Aⁱk) for all sets (i1, i₂, · · · , ik) of distinct subscripts chosen from (1, 2, · · · , n)

For example, for n = 3, we need

P (A₁A₂) = P (A₁)P (A₂), P (A₁A₃) = P (A₁)P (A₃), P (A2A3) = P (A2)P (A3)

and

P (A₁A₂A₃) = P (A₁)P (A₂)P (A₃)

Technically, we have defined “mutually independent” events, but we will shorten the name to “inde-pendent” to reduce confusion with “mutually exclusive.”

The definition of independence works two ways. If we can find P (A), P (B), and P (AB) then we can determine whether A and B are independent. Conversely, if we know (or assume) that A and B are independent, then we can use the definition as a rule of probability to calculate P (AB). Examples of each follow.

Example: Toss a die twice. Let A = {first toss is a 3} and B = {the total is 7}. Are A and B independent? (What do you think?) Using the definition to check, we get P (A) = ¹₆, P (B) = ₃₆⁶ (points (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) give a total of 7) and P (AB) = ₃₆¹ (only the point (3,4) makes AB occur).

Therefore, P (AB) = P (A)P (B) and so A and B are independent events.

Now suppose we change B to the event {total is 8}.

Then

P (A) = 1

6, P (B) = 5

36 and P (AB) = 1 36 Therefore P (AB)6= P (A)P (B)

and consequently A and B are dependent events.

This example often puzzles students. Why are they independent if B is a total of 7 but dependent for a total of 8? The key is that regardless of the first toss, there is always one number on the 2nd toss which makes the total 7. Since the probability of getting a total of 7 started off being ₃₆⁶ = ¹₆, the outcome of the 1st toss doesn’t affect the chances. However, for any total other than 7, the outcome of the 1st toss does affect the chances of getting that total (e.g., a first toss of 1 guarantees the total cannot be 8).

Example: A (pseudo) random number generator on the computer can give a sequence of independent random digits chosen from S = {0, 1, . . . , 9}. This means that (i) each digit has probability of ₁₀¹ of being any of 0, 1, . . . , 9, and (ii) the outcomes for the different trials are independent of one another.

We call this type of setting an “experiment with independent trials”. Determine the probability that (a) in a sequence of 5 trials, all the digits generated are odd

(b) the number 9 occurs for the first time on trial 10.

Solution:

(a) Define the events Ai = {digits from trial i is odd}, i = 1, . . . , 5.

Then

P (all digits are odd) = P (A₁A₂A₃A₄A₅)

= Q5 i=1

P (Ai),

since the A_i’s are mutually independent. Since P (A_i) = .5, we get P (all digits are odd) = .5⁵. (b) Define events A_i = {9 occurs on trial i}, i = 1, 2, . . . . Then we want

P ( ¯A₁A¯₂. . . ¯A₉A₁₀) = P ( ¯A₁)P ( ¯A₂) . . . P ( ¯A₉)P (A₁₀)

= (.9)⁹(.1),

because the Ai’s are independent, and P (Ai) = .1 = 1 − P ( ¯A_i).

Note: We have used the fact here that if A and B are independent events, then so are ¯A and B. To see this note that

P ( ¯AB) = P (B) − P (AB) (use a Venn diagram)

= P (B) − P (A)P (B) (since A and B are independent)

= (1 − P (A))P (B)

= P ( ¯A)P (B).

Note: We have implicitly assumed independence of events in some of our earlier probability calcula-tions. For example, suppose a coin is tossed 3 times, and we consider the sample space

S = {HHH, HHT, HT H, T HH, HT T, T HT, T T H, T T T } Assuming that the outcomes on the three tosses are independent, and that

P (H) = P (T ) = 1 2 on any single toss, we get that

P (HHH) = P (H)P (H)P (H) = (1 2)³= 1

8.

Similarly, all the other simple events have probability ¹₈. Note that in earlier calculations we assumed this was true without thinking directly about independence. However, it is clear that if somehow the 3 tosses were not independent then it might be a bad idea to assume each simple event had probability

1

8. (For example, instead of heads and tails, suppose H stands for “rain” and T stands for “no rain” on a given day; now consider 3 consecutive days. Would you want to assign a probability of ¹₈ to each of the 8 simple events?)

Note: The definition of independent events can thus be used either to check for independence or, if events are known to be independent, to calculate P (AB). Many problems are not obvious, and scientific study is needed to determine if two events are independent. For example, are the events A and B independent if, for a random child living in a country,

A = {live within 5 km. of a nuclear power plant}

B = {a child has leukemia}?

Such problems, which are of considerable importance, can be handled by methods in later statistics courses.

Problems:

4.3.1 A weighted die is such that P (1) = P (2) = P (3) = 0.1, P (4) = P (5) = 0.2, and P (6) = 0.3.

(a) If the die is thrown twice what is the probability the total is 9?

(b) If a die is thrown twice, and this process repeated 4 times, what is the probability the total will be 9 on exactly 1 of the 4 repetitions?

4.3.2 Suppose among UW students that 15% speaks French and 45% are women. Suppose also that 20% of the women speak French. A committee of 10 students is formed by randomly selecting from UW students. What is the probability there will be at least 1 woman and at least 1 French speaking student on the committee?

4.3.3 Prove that A and B are independent events if and only if A and B are independent.