Probability

Lesson

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).

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 |

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.

**
A coin is tossed twice.**

**Construct a tree diagram showing the results of the given experiment.****Use the tree diagram to find the probability of getting exactly 1 Head.****Use the tree diagram to find the probability of getting 2 heads.****Use the tree diagram to find the probability of getting no heads.****Use the tree diagram to find the probability of getting 1 Head and 1 Tail.**

**
Christa enters a competition in which she guesses the 3-digit code (from 000 to 999) which cracks open a vault containing one million dollars. If the 3-digit number to open the vault is randomly generated by a computer, what is the probability that it is:**

**an odd number?****an even number (including 000)?****a number greater than 123?****a number divisible by 10?****a number less than 321?**

**Two dice are rolled, and the numbers appearing on the uppermost faces are added. Find:**

**P(a total less than 6)****P(a total greater than or equal to 10)****P(a total of at least 7)**

Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal

Apply probability concepts in solving problems