Probability: Independence In-Depth & "At Least One" Problems


Check For Independence

Recall, two events are independent if the outcome of one event does not impact the outcome of the other event. That is, the two events are not related. There are three simple ways to check for independence using the probability rules that you have just learned:

  1. \(P(A) P(B)=P(A\;\cap\;B)\)
  2. \(P(A\mid B)=P(A)\)
  3. \(P(B\mid A)=P(B)\)

If any one of these statements is true, then all three statements are true and events A and B are independent.

 

Example:

P(G) = 0.8 and P(Kc) = 0.3. If P(G and K) = 0.6, are G and K independent?

Solution:

If G and K are independent, then P(G and K) = P(G)P(K). Before we can start, we must find P(K).

Recall that P(K) = 1 - P(Kc)  ⇒  P(K) = 1 - 0.3  ⇒  P(K) = 0.7

P(G)P(K) = (0.8)(0.7) = 0.56

Therefore, G and K are NOT independent because P(G)P(K) \(\ne\) 0.6 


Independent vs. Disjoint Events

Note that disjoint and independent events are different. Events are considered disjoint or mutually exclusive if they cannot occur at the same time. For example, being early to class and being late to the same class would be considered disjoint events because an individual cannot be labeled as both. Now, if someone told you that a student was early to class, does that change the probability that this same student will be late to class? The answer, of course, is yes. Thus, disjoint/mutually exclusive events can never be labeled as independent events.

 

 


"At Least One" Problems

Often, people find themselves wanting to find the probability of "at least one". For example:

  • A female student applies to 10 colleges that have similar criteria for acceptance, what is the probability that at least one will accept her?
  • A guy asks 10 different girls to go to the prom (all at the same time!), what is the probability that at least one will say yes?

Let's take the first example to better understand this type of problem. One approach for the female student would be to calculate the P(X =1) + P(X = 2) + ... + P(X = 10). This approach, however, is very cumbersome. If all the probabilities are equal, however, another approach would be to recognize the following as true: P(at least one college accepts) + P(none accept) = 1. Once this relationship is understood, then the next step is to simply recognize that for all "at least 1" problems, the following formula applies:

P(at least one) = 1 - P(none)

 

Examples:

1. A fair die is rolled 10 times. Find the probability that there is at least one 5. 

Solution:

P(at least one 5) = 1 - P(none of the rolls equal 5)
                                                       = 1 - P(all 10 rolls are not five)      ⇒     P(5) = 1/6   P(not 5) = 5/6
                                                       = 1 - (5/6)10
                                                       = 0.838

 

2. A man purchases a box of 75 decorative light bulbs for his restaurant. If the probability that any light bulb is defective is 0.02, calculate the probability that at least one light bulb is defective in the box of 75. 

Solution:

P(at least 1 light bulb is defective) = 1 - P(none of the light bulbs are defective)
                                                       = 1 - P(all 75 light bulbs are good)      ⇒     P(bad light bulb) = 0.02    P(good light bulb) = 0.98
                                                       = 1 - (0.98)75
                                                       = 0.7802