Independent events

Independent Events

We say that two events are independent if the occurrence of one event does not affect the probability of the other occurring.  

For example,

• Tossing a coin is an independent event.  It really doesn't matter what the last 100 tosses have been, the next toss has a 50% chance of being a head and 50% chance of being a tail.  In fact any type of event that operates with replacement, or consecutive equally likely events is independent.
• Tossing a coin and then rolling a die are independent events, because they use completely different objects.  The die is not affected by the coin and vice versa.  Any type of events that use different objects are independent.  

To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule is defined symbolically as

$\text{P(A and B)}$P(A and B)$=$=$P(A$P(A$B)$B)$=$=$P(A)$P(A) x $P(B)$P(B), for independent events. The symbol that looks like a shoehorn is used in the case where we have one event AND another occurring.

This means you can also test for independence by verifying if P(A) x P(B)  = P(A and B).

 

Examples

Question 1

A coin is tossed and a single 6-sided die is rolled. Find the probability of flipping a tail on the coin and rolling a 4 on the die.

$\text{P(Tail) }=\frac{1}{2}$P(Tail) =12​

$\text{P(4) }=\frac{1}{6}$P(4) =16​

$\text{P(Tail and 4) }$P(Tail and 4)	$=$=	$\text{P(Tail) }\times\text{P(4) }$P(Tail) ×P(4)
	$=$=	$\frac{1}{2}\times\frac{1}{6}$12​×16​
	$=$=	$\frac{1}{12}$112​

 

Question 2 (LARGE SAMPLE SPACE)

A nationwide survey found that $64%$64% of people in a small country town have unreliable internet access.  If 3 people are selected at random, what is the probability that all three have unreliable internet access?

P(I) x P(I) x P(I) = $0.64\times0.64\times0.64=0.262144$0.64×0.64×0.64=0.262144 

Now why are the events independent here? You may think that in the case of selecting people from a population, we should not replace the first person before selecting the next. This would make the selections dependent. But consider the following:

Say the population is $1000000$1000000, and $640000$640000 of them do not have reliable internet access.

P(first person has unreliable internet access) = $\frac{640000}{1000000}=0.64$6400001000000​=0.64

If we remove one of these people from the population before the second draw, there would be $999999$999999 people left in the population, and $639999$639999 of them would have unreliable internet access:

P(second person has unreliable internet access) = $\frac{639999}{999999}=0.639$639999999999​=0.639..

Without going any further, you can see that the probability does not change that much.

So for a large sample space, the probability changes so little that we can consider successive events as being independent.

Examples

Question 1

Question 2

Question 3