Elementary Statistics and Computer Application

Statistical Probability, Axioms for Probability & Conditional Probability

Statistical (or) Empirical Probability (or) a-posteriori Probability

• If an experiment is repeated a number (n) of times, an event ‘A’ happens ‘m’ times then the statistical probability of ‘A’ is given by

Axioms for Probability

• The probability of an event ranges from 0 to 1. If the event cannot take place its probability shall be ‘0’ if it certain, its probability shall be ‘1’.
• Let E1, E2, …., En be any events, then P(Ei) 0.
• The probability of the entire sample space is ‘1’. (i.e.) P(S) = 1.
Total Probability,

• If A and B are mutually exclusive (or) disjoint events then the probability of occurrence of either A (or) B denoted by P(AUB) shall be given by
P(AUB) = P(A) + P(B)
P(E1UE2U….UEn) = P(E1) + P(E2) + …… + P(En)
If E1, E2, …., En are mutually exclusive events.

Example 1: Two dice are tossed. What is the probability of getting (i) Sum 6 (ii) Sum 9?
Solution:

• When 2 dice are tossed. The exhaustive number of cases is 36 ways.
• (i) Sum 6 = { (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) }
• .^.. Favourable number of cases = 5
• P(Sum 6) =
• (ii) Sum 9 = { (3, 6), (4, 5), (5, 4), (6, 3) }
• .^.. Favourable number of cases = 4
 
P(Sum 9) = =

Example 2: A card is drawn from a pack of cards. What is a probability of getting (i) a king (ii) a spade (iii) a red card (iv) a numbered card?

Solution

• There are 52 cards in a pack.
• One can be selected in 52C1 ways.
• .^.. Exhaustive number of cases is = 52C1 = 52.
(i) A king

• There are 4 kings in a pack.
• One king can be selected in 4C1 ways.
• .^..Favourable number of cases is = 4C1 = 4
• Hence the probability of getting a king =
(ii) A spade

• There are 13 kings in a pack.
• One spade can be selected in 13C1 ways.
• .^..Favourable number of cases is = 13C1 = 13
• Hence the probability of getting a spade =
(iii) A red card

• There are 26 kings in a pack.
• One red card can be selected in 26C1 ways.
• .^.. Favourable number of cases is = 26C1 = 26
• Hence the probability of getting a red card =
(iv) A numbered card

• There are 36 kings in a pack.
• One numbered card can be selected in 36C1 ways.
• .^.. Favourable number of cases is = 36C1 = 36
• Hence the probability of getting a numbered card =

Example 3: What is the probability of getting 53 Sundays when a leap year selected at random?
Solution:

• A leap year consists of 366 days.
• This has 52 full weeks and 2 days remained.
• The remaining 2 days have the following possibilities.
• (i) Sun. Mon (ii) Mon, Tues (iii) Tues, Wed (iv) Wed, Thurs (v) Thurs, Fri (vi) Fri, Sat (vii) Sat, Sun.
• In order that a lap year selected at random should contain 53 Sundays, one of the 2 over days must be Sunday.
• .^.. Exhaustive number of cases is = 7
• .^.. Favourable number of cases is = 2
• .^.. Required Probability is =

Conditional Probability

• Two events A and B are said to be dependent, when B can occur only when A is known to have occurred (or vice versa). The probability attached to such an event is called the conditional probability and is denoted by P(A/B) (read it as: A given B) or, in other words, probability of A given that B has occurred.


• If two events A and B are dependent, then the conditional probability of B given A is,