document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164741>Question 1164741</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A trading company has eight computers that it uses to trade on the New York Stock Exchange(NYSE). The probability of a computer failing in a day is 0.005, and the computers fail independently. Computers are repaired in the evening and each day is an independent trial.\r<BR>\n" );
document.write( "\n" );
document.write( "(a) What is the probability that all eight computers fail in a day?<BR>\n" );
document.write( "(b) What is the mean number of days until a specific computer fails?<BR>\n" );
document.write( "(c) What is the mean number of days until all eight computers fail in the same day\r<BR>\n" );
document.write( "\n" );
document.write( "compute the solution in R ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #853520 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=853520\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This problem involves two different probability distributions: the **Binomial Distribution** (for the number of computers failing in a single day) and the **Geometric Distribution** (for the number of days until a specific event occurs).\r<BR>\n" );
document.write( "\n" );
document.write( "### (a) Probability that all eight computers fail in a day\r<BR>\n" );
document.write( "\n" );
document.write( "Since the computers fail independently and each has the same probability of failure, we use the Binomial probability formula .\r<BR>\n" );
document.write( "\n" );
document.write( "For \"all eight fail,\" we have  and :\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "### (b) Mean number of days until a specific computer fails\r<BR>\n" );
document.write( "\n" );
document.write( "The number of trials (days) until the first \"success\" (failure of the computer) follows a Geometric Distribution. The mean (expected value) of a geometric distribution is .\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "### (c) Mean number of days until all eight computers fail on the same day\r<BR>\n" );
document.write( "\n" );
document.write( "This is another Geometric Distribution problem, but the \"success\" event is now the extremely rare occurrence of all eight computers failing simultaneously.\r<BR>\n" );
document.write( "\n" );
document.write( "1. First, find the probability of the event  calculated in part (a).<BR>\n" );
document.write( "2. The mean days until this occurs is .\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "### R Code Solution\r<BR>\n" );
document.write( "\n" );
document.write( "You can run the following code in R to get the exact values:\r<BR>\n" );
document.write( "\n" );
document.write( "```r<BR>\n" );
document.write( "# Parameters<BR>\n" );
document.write( "n_computers <- 8<BR>\n" );
document.write( "p_fail <- 0.005\r<BR>\n" );
document.write( "\n" );
document.write( "# (a) Probability all 8 fail in a single day<BR>\n" );
document.write( "# We use dbinom(k, size, prob)<BR>\n" );
document.write( "prob_all_fail_day <- dbinom(8, size = n_computers, prob = p_fail)<BR>\n" );
document.write( "# Or simply p_fail^8<BR>\n" );
document.write( "# prob_all_fail_day <- p_fail^8\r<BR>\n" );
document.write( "\n" );
document.write( "print(paste(\"Prob all 8 fail in a day:\", format(prob_all_fail_day, scientific = TRUE)))\r<BR>\n" );
document.write( "\n" );
document.write( "# (b) Mean days until a specific computer fails<BR>\n" );
document.write( "# E[X] = 1/p<BR>\n" );
document.write( "mean_days_specific <- 1 / p_fail<BR>\n" );
document.write( "print(paste(\"Mean days until a specific computer fails:\", mean_days_specific))\r<BR>\n" );
document.write( "\n" );
document.write( "# (c) Mean days until all 8 fail in the same day<BR>\n" );
document.write( "# E[X] = 1 / P(all 8 fail)<BR>\n" );
document.write( "mean_days_all_fail <- 1 / prob_all_fail_day<BR>\n" );
document.write( "print(paste(\"Mean days until all 8 fail in the same day:\", format(mean_days_all_fail, big.mark = \",\")))\r<BR>\n" );
document.write( "\n" );
document.write( "```\r<BR>\n" );
document.write( "\n" );
document.write( "### Interpretation of Results\r<BR>\n" );
document.write( "\n" );
document.write( "* **Part (a):** The probability is . This is infinitesimally small, meaning it is almost impossible for this to happen on any given day.<BR>\n" );
document.write( "* **Part (b):** You can expect a specific computer to fail roughly once every **200 days**.<BR>\n" );
document.write( "* **Part (c):** You can expect all eight computers to fail on the same day once every ** days**. For context, this is significantly longer than the age of the universe.\r<BR>\n" );
document.write( "\n" );
document.write( "Would you like me to calculate the probability that **at least one** computer fails in a day? </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );