# SOLUTION: 1) a boy and a girl have to toss a coin . The winner is who gets the head . If the girl starts first , what's the probability the boy wins ? 2) In a factory three new machines A

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: 1) a boy and a girl have to toss a coin . The winner is who gets the head . If the girl starts first , what's the probability the boy wins ? 2) In a factory three new machines A      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth

 Question 409490: 1) a boy and a girl have to toss a coin . The winner is who gets the head . If the girl starts first , what's the probability the boy wins ? 2) In a factory three new machines A,B,C have the probability of producing faulty lamps as 0.9 , 0.7 and 0.6 respectively . If three lamps are selected randomly find the probabilty that at least two lamps are produced by the machine B ? 3) Three new computers have the probability of failure as 0.75 , 0.65 and 0.5 . What's the probabilty that two computers fail in the first year ?Answer by sudhanshu_kmr(1152)   (Show Source): You can put this solution on YOUR website!1) As here the girl starts first, she can win on first, third, fifth ... toss probability of win on first chance = 1/2 probability that she can win on second chance(i.e third toss) = 1/2*1/2*1/2 (for win on second chance outcome should be Tail, Tail then Head ) similarly probability on third chance = (1/2)^5 probability the girl win P(A)= 1/2 + (1/2)^3 + (1/2)^5 .............to infinity { here sum of geometric progression , a = 1/2 and r = 1/4 } = 1/2 /[1 -1/4] = 2/3 probability the boy win = P(A') = 1-P(A) = 1/3 2) three lamps can be faulty by following ways... A B C Probability 1 1 1 - 9/10 * 7/10* 6/10 = 378/1000 2 1 0 - 9/10 * 9/10 * 7/10 = 567/1000 2 0 1 - 9/10 * 9/10 * 6/10 = 486/1000 3 0 0 - 9/10 * 9/10 * 9/10 = 729/1000 1 2 0 - 441/1000 ( favorable condition) 0 2 1 - 294/1000 ( favourable condition) 0 3 0 - 343/1000 ( favourable condition) 1 0 2 - 324/1000 0 1 2- 252/1000 0 0 3- 216/1000 using Bye's theory P(a) = probability of favorable condition/ probability of total condition = [ 441/1000 + 294/ 1000+ 343/1000]/ [ 378/1000 + .......+ 216/1000] = 1078 / 4030 3) two computers may fail by following ways... probability : A and B not C = 75/100 * 65/100 * (1- 50/100) A and C not B = 75/100 * 50/100 * 35/100 B and C not A = 65/100 * 50/100 * 25/100 probability that two computers fail = [243750 +131250 + 81250 ] /1000000 = 456250/1000000 = 0.456250