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
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