# Probability formula

## What is probability formula

What is probability formula If a random experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes favours to A, then the probability of event A, P (A) =

i.e.

Remarks:

• If the probability of certain event is one, it doesn’t mean that event is going to happen with certainty! Infact we would be just predicting that, the event is most likely to occur in comparison to other events. Predictions depend upon the past information and of course also on the way of analysing the information at hand!!

• Similarly if the probability of certain event is zero, it doesn’t mean that, the event can never occur!

Example -1:    In a single cast with two fair dice, what is the chance of throwing

(i) two 4′s

(ii) a doublet

(iii) five - six

(iv) a sum of 7

Sol:     (i) There are 6 × 6 equally likely cases ( as any face of any die may turn up)

⇒ 36 possible outcomes. For this event, only one outcome (4-4) is favourable

∴ probability = 1/36).

(ii)A doublet can occur in six ways {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.

Therefore probability of doublet = 6/36 = 1/6.

(iii) Two favourable outcomes {(5, 6), (6, 5)}

Therefore probability = 2/36 = 1/18

(iv) A sum of 7 can occur in the following cases {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} which are 6 in number. Therefore probability = 6/36 = 1/6

Example -2:    From a bag containing 5 white, 7 red and 4 black balls a man draws 3 balls  at random, find the probability of being all white.

Sol:    Total number of balls in the bag  =5+7+4= 16

The total number of ways in which 3 balls can be drawn is

16C3 = = 560

Thus sample space S  for this experiment has 560 outcomes  i.e. n(S) =560

Let E be the event of all the three balls being white. Total number of white balls is 5. So the number of ways in which 3 white balls can be drawn = 5C3= =10

Thus E has 10 elements of S, ∴ n(E) = 10

∴ Probability of E, P(E) = ==

Example -3:    From a set of 17 cards 1, 2, 3, ……16, 17, one is drawn at random. Find the probability that number on the drawn card would be divisible by 3 or 7.

Sol:    Numbers which are divisible by three are 3, 6, 9, 12, 15.  Similarly numbers which are divisible by 7 are 7, 14.

⇒  Probability of the number written on the card to be divisible by 3 or 7 is

= .

### Mutually Exclusive Events

A set of events is said to be mutually exclusive if the occurrence of one of them precludes the occurrence of any of the other events. For instance, when a pair of dice is tossed, the events a sum of 4 occurs’, ‘a sum of 10 occurs’ and ‘a sum of 12 occurs’ are mutually exclusive. Simply speaking, if two events are mutually exclusive they can not occur simultaneously. Using set theoretic notation, if A1, A2,.. An be the set of mutually events then Ai ∩ Aj =φ for i ≠j and 1≤ i, j≤ n.

### Independent Events

Events are said to be independent if the occurrence or non-occurrence of one does not affect the occurrence or non-occurrence of other. For instance, when a pair of dice is tossed, the events ‘first die shows an even number’ and ‘second die shows an odd number’ are independent. As the outcome of second die does not affect the outcome of first die. It should also be noted that these two events are not mutually exclusive as they can occur together.

Remarks:

• Distinction between mutually exclusive and independent events should be clearly made. To be precise, concept of mutually exclusive events is set theoretic in nature while the concept of independent events is probabilistic in nature.
• If two events A and B are mutually exclusive, they would be strongly dependent as the occurrence of one precludes the occurrence of the other.

### Exhaustive Event

A set of events is said to be exhaustive if the performance of random experiment always result in the occurrence of atleast one of them. For instance, consider a ordinary pack of cards then the events ‘drawn card is heart’, drawn card is diamond’, ‘drawn card is club’ and ‘drawn card is spade’ is set of exhaustive event. In other words all sample points put together (i.e. sample space itself) would give us an exhaustive event.

If ‘E’ be an exhaustive event then P(E) = 1.