Question 293293
A fair coin is tossed until first tail comes up. Let A be the event that the number of tosses is divisible by 5, 

Find P(A)
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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) = {{{(1/2)^5}}}, which means the 5 tosses went HHHHT 
P(x=10) = {{{(1/2)^10}}} which means the 10 tosses went HHHHHHHHHT
P(x=15) = {{{(1/2)^15}}} which means the 15 tosses went HHHHHHHHHHHHHHT
...
P(x=5n) = {{{(1/2)^(5n)}}}
...

P(A) = the sum of an infinite geometric series with {{{a[1]=1/(2^5)=1/32}}}
and {{{r=(1/2)^5=1/32}}}

{{{S[infinity]=a[1]/(1-r)=(1/32)/(1-1/32)=1/(32-1)=1/31}}}

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Let B be the event that the number of tosses is less than 6 
Find P(B)
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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) = {{{(1/2) + (1/2)^2 + (1/2)^3 + (1/2)^4 + (1/2)^5}}} 

P(B) = {{{1/2+1/4+1/8+1/16+1/32=16/32+8/32+4/32 +2/32 + 1/32=31/32}}}

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Find P(AuB) 
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Use the formula:

P(A <font face = "arial">U</font> B) = P(A or B) = P(A) + P(B) - P(A and B)

We have P(A) = {{{1/31}}},  P(B) = {{{31/32}}}

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 {{{(1/2)^5=1/32}}}=P(A and B)

P(A <font face = "arial">U</font> B) = P(A or B) = P(A) + P(B) - P(A and B)

P(A <font face = "arial">U</font> B) = P(A or B) = {{{1/32 + 31/32 - 1/32=31/32}}}

Edwin</pre>