Ask Question
21 May, 03:55

A box contains four 75 W lightbulbs, three 60 W lightbulbs, and three burned-out lightbulbs. Two bulbs are selected at random from the box without replacement. Let X represent the number of 75 W bulbs selected. Find the probability mass function for X. Show that X follows a valid probability mass function.

a. Find P (X > 0)

b. Find μx

c. Find σx^2

+4
Answers (1)
  1. 21 May, 03:58
    0
    a. 0.689

    b. 0.8

    c. 0.427

    Step-by-step explanation:

    The given scenario indicates hyper-geometric experiment because because successive trials are dependent and probability of success changes on each trial.

    The probability mass function for hyper-geometric distribution is

    P (X=x) = kCx (N-k) C (n-x) / NCn

    where N=4+3+3=10

    n=2

    k=4

    a.

    P (X>0) = 1-P (X=0)

    The probability mass function for hyper-geometric distribution is

    P (X=x) = kCx (N-k) C (n-x) / NCn

    P (X=0) = 4C0 (6C2) / 10C2=15/45=0.311

    P (X>0) = 1-P (X=0) = 1-0.311=0.689

    P (X>0) = 0.689

    b.

    The mean of hyper-geometric distribution is

    μx=nk/N

    μx=2*4/10=8/10=0.8

    c.

    The variance of hyper-geometric distribution is

    σx²=nk (N-k). (N-n) / N² (N-1)

    σx²=2*4 (10-4). (10-2) / 10²*9

    σx²=8*6*8/900=384/900=0.427
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A box contains four 75 W lightbulbs, three 60 W lightbulbs, and three burned-out lightbulbs. Two bulbs are selected at random from the box ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers