Hyper-geometric distribution:

    Suppose a population consists of N items, k of which are successes. A random sample drawn from that population consists of n items, x of which are successes.


    Then the hyper-geometric probability is:

    h(x, N, n, k) = [ kCx][(n-k)C(n-k)]/[NCn]

    Where,
    N -Number of items in the population.

    k: Number of items in the population that are classified as successes.

    n: Number of items in the sample.

    x: Number of items in the sample that are classified as successes.

  1. 10 cards are selected randomly from a deck of cards without replacements. What is the probability of getting exactly 5 black cards?

    Solution:

    • N = 52, k = 26; since there are 26 black cards in a deck.
    • n = 10; since we randomly select 10 cards from the deck.
    • x = 5; since 5 of the cards we select are black.
    • h(x; N, n, k) = [ kCx][(n-k)C(n-k)]/[NCn]
    • h(5; 52, 10, 26) = [26C5*26C5]/[52C10]
    • h(5; 52, 10, 26) = [ 65780 * 65780] / [ 15820024220 ] = 0.2735


  2. There are 20 black marbles and 20 white marbles out of which 10 marbles are being chosen. Find the probability that there are 4 white marbles in them.

    Solution:

    • N = 40, k = 20; since there are 20 white marbles.
    • n = 10; randomly select 10 cards from the deck.
    • x = 4; since 4 of the white marbles .
    • h(x; N, n, k) = [kCx][(n-k)C(n-k)]/[NCn]
    • h(4; 40, 10, 20) = [20C4*20C6]/[40C10]
    • h(4; 40, 10, 20) = [ 4845 * 38760] / [ 847660528 ] = 0.2215


  3. Out of 150 aspirants qualifying an exam, 25 were drawn randomly. If 30 out of 150 qualified aspirants are female, then find the probability that 8 out of 25 chosen are females.


  4. Solution:

    • N = 150, k = 30;
    • n = 25
    • x = 8;
    • h(x; N, n, k) = [kCx][(N-k)C(n-x)]/[NCn]
    • h(8; 150, 25, 30) = [30C8*120C17]/[150C25]
    • h(8; 150, 25, 30) = 0.0568


  5. There are 66 bulbs in a house out which 33 are defective. If 22 bulbs are picked randomly, find the probability that at least one is defective.


  6. Solution:

    • h(x; N, n, k) = [kCx][(N-k)C(n-x)]/[NCn]
    • P(X=0) = [3C0*3C2]/[6C2]
    • = 0.2
    • = 0.2
    • Probability of at least one defective bulb = 1 – p(X=0) = 1 – 0.2
    • = 0.8