Example 1

The probability of a student getting an A in this course is 0.25 (Not True!) and the probability of getting a B is 0.30 (again Not True!). What is the probability of getting an A or a B? According to Rule 4.a , P(A or B) = P(A) + P(B) = 0.25 + 0.30 = 0.55. In this example events A and B are mutually exclusive since a student cannot get both an A and a B for a course, but instead can only get one grade. Take a look at the Venn diagram example below:

Example 2

The probability of a student in this course getting an A is 0.25; the probability of being female is 0.60. What is the probability of getting an A or being female? .......Can't say? Well, we could add probabilities in example 1 since the two events were mutually exclusive, but in this example getting an A and being female may both occur. So we just cannot simply add the probability of one event to the probability of the other event - if we did we would be counting twice the probability where the two events overlap, A and Female. To solve this problem then we need to know the probability of both A and Female occurring - P(A and Female). Suppose, for example, this probability were 0.15. We would find P(A and Female) by Rule 4.b : P(A and Female) = P(A) + P(Female) - P(A and Female) = 0.25 + 0.60 - 0.15 = 0.70. Using the Venn diagram below might help to explain this how this works:

Example 3

Now, pick a playing card at random for four cards, one each a diamond, club, spade, and heart. Replace the card and pick again. What is the probability that the first card is a heart and the second is heart? That is, find P(Heart and Heart).

(If you don't know what's in a standard deck of playing cards then follow this link...)

The answer: 1/4 × 1/4 = 1/16, (see Rule 5.b )

Example 4

Repeat example 3, but do NOT replace the card. Now what is P(Heart and Heart). The answer is zero since once you draw the heart on your first try and don't replace the card there is not another heart to select.

In Example 3, because the first card is replaced prior to selecting the second card, the two events are independent - neither draw influences the other. In Example 4, because the first card is not replaced the selections are dependent.

Example 5

Roll a die 4 times. What is the probability of getting at least one 2? If we let A (or any letter for that matter) represent the event "number of 2's achieved when rolling a die 4 times", then we are searching for P(A = 1), and this probability is:

= probability of 1^st two on the 1^st roll or 2^nd roll or 3^rd roll or 4^th roll

= probability of 1^st two on 1^st roll + . . . + probability of 1^st two on 4^th roll. {Rule 4.2}

P(1^st two on 1^st roll) = 1/6

P(1^st two on 2^nd roll) = P(no two 1^st roll) * P(1^st two on 2^nd roll) = 5/6 × 1/6 = 5/36

{note that Rule 3 was used for P(no two 1^st roll)}

P(1^st two on 3^rd roll) = P(no two 1^st roll) × P(no two 2^nd roll) × P(1^st two on 3^rd roll) = 5/6 × 5/6 × 1/6 = 25/216

P(1^st two on 4^th roll) = P(no two 1^st roll) × P(no two 2^nd roll) × P(no two 3^rd roll) × P(two on 4^th roll) = 5/6 × 5/6 × 5/6 × 1/6 = 125/1296

So now that we have all of these probabilities we add them to get our answer:

P(getting at least one 2 on four die rolls) = 1/6 + 5/36 + 25/216 + 125/1296 = 671/1296 = 0.52

WHEW!!!! Is there an easier way, please!? Yes there is!

The easier solution is to apply Rule 3 regarding the complement. By Rule 3 , the probability of getting at least one 2 in 4 rolls is equal to (1 - the probability of not getting any 2's in 4 rolls). That is, the earlier notation of using P(A = 1) = 1 - P(A < 1) and P(A < 1) only leaves 0. So this expression becomes 1 - P(A = 0). Since the probability of not getting a 2 on any roll of a die is 5/6, the probability of not getting a 2 on 4 rolls of die equals 5/6 × 5/6 × 5/6 × 5/6 = 625/1296.

And, therefore, 1 - 625/1296 = 671/1296 which is precisely what we received before!

Much easier!