SOLUTION: A fair coin is tossed until first tail comes up. Let A be the event that the number of tosses is divisible by 5, b be the event that the number of tosses is less than 6 Find P(A

The other tutor's solution is incorrect.

P(A) = P(x=5 or x=10 or x=15 or ... x=5n or ...)

Since "or" implies "add"

P(A) = P(x=5) + P(x=10) + P(x=15) + ... + P(x=5n) + ...

P(x=5) = , which means the 5 tosses went HHHHT 
P(x=10) =  which means the 10 tosses went HHHHHHHHHT
P(x=15) =  which means the 15 tosses went HHHHHHHHHHHHHHT
...
P(x=5n) = 
...

P(A) = the sum of an infinite geometric series with 
and 



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P(B) = P(T or HT or HHT or HHHT or HHHHT or HHHHT) =

Since "or" implies "add"

P(B) = P(T) + P(HT) + P(HHT) + P(HHHT) + P(HHHHT) =

P(B) =  

P(B) = 

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Use the formula:

P(A U B) = P(A or B) = P(A) + P(B) - P(A and B)

We have P(A) = ,  P(B) = 

The event "A and B" is the event HHHHT, the only event
which would be both a multiple of 5 tosses and also less
than 6 tosses.

HHHHT has probability =P(A and B)

P(A U B) = P(A or B) = P(A) + P(B) - P(A and B)

P(A U B) = P(A or B) = 

Edwin