Statistical Probability, Axioms for Probability & Conditional Probability

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
    Statistical
    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) symbol 0.
    • The probability of the entire sample space is ‘1’. (i.e.) P(S) = 1.
    Total Probability, 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) = Ans
    • (ii) Sum 9 = { (3, 6), (4, 5), (5, 4), (6, 3) }
    • ... Favourable number of cases = 4
       
    P(Sum 9) =ANS = ANS

    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 = ANS
    (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 =ANS
    (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 = ANS
    (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 = ANS

    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 = ANS

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

Last modified: Friday, 16 March 2012, 7:47 PM