Probability of “not”, “and” and “or” events

If the events A and B are not mutually exclusive,

 the probability is: P(A or B) = P(A) + P(B) – P(A and B).

and

P(not A) = 1 – P(A)

Either/or probability refers to the probability that one event or the other will occur. For example, what is the probability that you will draw a Jack or a three from a normal deck of cards? Or, what is the probability that you will roll a 3 or a 5 when rolling a normal 6-sided die? To solve this type of probability problem, here is the formula you will use:

P(A or B) = P(A) + P(B)

To find the probability of each event, simply divide the amount of favorable events by the amount of total events. A favorable event is an event that you want to occur. In the earlier card question, the favorable event is drawing either a Jack or a three. The total number of events is the total number of things that could occur, whether favorable or not.

So, to continue on and solve this card drawing question, we have determined that A is the probability of drawing a Jack, and B is the probability of drawing a three.

There are 4 Jacks in a normal deck of cards, so the number of favorable events (drawing a Jack) is 4. The total number of events is 52 since there are 52 cards in a deck of cards. This means that the probability of drawing a Jack is 4/52, which can be reduced to 1/13.

P(B), or the probability of drawing a three, is also 1/13 because there are 4 threes in a deck of cards and, as before, there are 52 total cards in the deck.

To finish answering the question and find the probability of drawing either a Jack or a three, we use the equation P(A or B) = P(A) + P(B). P(A or B) is equal to 1/13 + 1/13, which is 2/13

To solve the dice question mentioned earlier, follow the same steps. P(A), or the probability of rolling a 3, is 1/6. There is one 3 (the favorable event) and 6 sides on the die (the total events).

P(B) is the probability of rolling a 5 and it's the same, 1/6. Therefore, the probability of rolling either a 3 or a 5 is P(A or B) is equal to 1/6 + 1/6, which is 2/6, or 1/3.

These events are called non-overlapping events, or events that are independent of each other. There are also overlapping events, which are events that are not independent of each other.

Example :  A die has sides numbered 1-9. You roll the die. What is the probability that the number you obtain is odd or prime.

Solution: Let A be the event that the number is odd and B the event that the number is a prime.

Then A and B are overlapping events 

since a number can be both prime and odd from among the numbers 1-9.

The odd numbers in 1-9 are 1, 3, 5, 7 and 9 (total of five), while the prime numbers in 1-9 are 1, 3, 5 and 7 (total of four). Hence the numbers in 1-9 that are odd AND prime are 1, 3, 5 and 7 (total of four).

Hence:

P(A or B) = P(A) + P(B) - P(A and B)

Example : A box has 6 red marbles, 7 blue marbles and 8 green marbles. You draw one marble at random from the box. What is the probability that the marble is red or green?

Solution:

Let A be the event that the marble is red and B be the event that the marble is green.

Since A and B are not overlapping, that marble cannot be both red AND green. So we get the probability as follows:

P(A or B) = P(A) + P(B)

Example : What is the probability that a card taken from a standard deck, is an Ace?

Solution:

Total number of cards a standard pack contains = 52

Number of Ace cards in a deck of cards = 4

So, the number of favourable outcomes = 4

Now, by looking at the formula,

Probability of selecting an ace from a deck is,

P(Ace) = (Number of favourable outcomes) / (Total number of favourable outcomes)

P(Ace) = 4/52

= 1/13

So we can say that the probability of getting an ace is 1/13.

Example : Calculate the probability of getting an odd number if a dice is rolled.

Solution:

Sample space (S) = {1, 2, 3, 4, 5, 6}

n(S) = 6

Let “E” be the event of getting an odd number, E = {1, 3, 5}

n(E) = 3

So, the Probability of getting an odd number is:

P(E) = (Number of outcomes favorable)/(Total number of outcomes)

= n(E)/n(S)

= 3/6

= ½

Example : Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that

(a) Both Anil and Ashima will not qualify the examination.

(b) Atleast one of them will not qualify the examination and

(c) Only one of them will qualify the examination.

Solution: Let E and F denote the events that Anil and Ashima will qualify the examination,

respectively.

Given that P(E) = 0.05, P(F) = 0.10 and P(E Ç F) = 0.02.

Then

  1. The event ‘both Anil and Ashima will not qualify the examination’ may be expressed as

Ç F´.

Since, E´ is ‘not E’, i.e., Anil will not qualify the examination and F´ is ‘not F’,

i.e., Ashima will not qualify the examination.

Also E´ Ç F´ = (E È F)´ (by Demorgan's Law)

Now P(E È F) = P(E) + P(F) – P(E Ç F)

or P(E È F) = 0.05 + 0.10 – 0.02 = 0.13

Therefore P(E´ Ç F´) = P(E È F)´ = 1 – P(E È F) = 1 – 0.13 = 0.87

(b) P (atleast one of them will not qualify)

= 1 – P(both of them will qualify)

= 1 – 0.02 = 0.98

(c) The event only one of them will qualify the examination is same as the event either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima will qualify)

i.e., E Ç F´  or  E´ Ç F,  where E Ç F´ and E´ Ç F are mutually exclusive.

Therefore, P(only one of them will qualify) = P(E Ç F´ or E´ Ç F)

=  P(E Ç F´) + P(E´ Ç F) = P (E) – P(E Ç F) + P(F) – P (E Ç F)

=  0.05 – 0.02 + 0.10 – 0.02 = 0.11

Example : If A, B, C are three events associated with a random experiment,

prove that

P(AÈBÈC) = P(A) + P(B) + P(C) - P(AÇB) - P(AÇC) – P ( B Ç C) + P ( A Ç B Ç C)

Solution: Consider E = B È C

so that

P (A È B È C ) = P (A È E )

= P(A) + P(E) - P(A ÇE) ............... (1)

Now

P(E) = P(BÈC) = P(B) + P(C) − P(BÇC) ... ……. (2)

Also AÇE = AÇ(BÈC) = (A ÇB)È(AÇC) [using distribution property of

intersection of sets over the union].

Thus

P(A ÇE) = P(A ÇB) + P(A ÇC)– P [(AÇB)Ç(AÇC)]

= P(AÇ B) + P(A ÇC) – P[A ÇBÇC] .................... (3)

Using (2) and (3) in (1), we get

P[AÈ BÈC] = P(A) + P(B) + P(C) - P(BÇC)– P(AÇ B) - P(A ÇC) + P(A ÇBÇC)