Question 409490
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