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

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, b be the event that the number of tosses is less than 6
Find P(A) P(B) P(AuB)
Can someone please talk me through the method to solve these
Thank You
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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) = P(5) + P(10) + P(15)+ ...
= (1/32) + (1/2)^10 + (1/2)^15...
-----------------------------------
P(B) = P(1)+P(2)+...+P(5)
= (1/2) + (1/2)^2 + ...+(1/2)^5
-----------------------------------
P(AUB) = (1/2) + (1/2)^2 +...(1/2)^5 + (1/2)^10 + (1/2)^15 + ...
================================================================
Cheers,
Stan H.
====================

Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!
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)

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 ------------------------

Let B be the event that the number of tosses is less than 6
Find P(B)

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) = ---------------------

Find P(AuB)

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

RELATED QUESTIONS
A coin is tossed continuously until either a head turns up or a total of three tails... (answered by stanbon)
A fair coin is tossed three times. Let X be the number of heads in the tosses minus the... (answered by Edwin McCravy)
. A fair die is tossed 8 times. Determine the probabilities of the following events: a.... (answered by greenestamps)
A fair coin is tossed n times. Let X be the number of times that a head occurs in the n... (answered by CPhill)
Determine whether the events in the following situations are independent: a. In a normal (answered by Boreal)
A fair coin is tossed, and a fair die is thrown. Write down sample spaces for (a) the... (answered by Edwin McCravy)
a coin is tossed until for the first time the same result occurs in twice succession.Find (answered by Edwin McCravy)
A balanced die is tossed twice. If A=Event (an even number comes up on the first toss), B (answered by CPhill)
A fair coin is tossed 4 times. Let x = number of tails observed in the 4 tosses. Find... (answered by Edwin McCravy)