A coin is tossed repeatedly until it falls heads for the fou

A coin is tossed repeatedly until it falls heads for the fourth time. what is the probability that the fourth head occurs on the seventh toss?

If nine coins are tossed what is the probability that the number of heads in even? how about 99 coins? 100 coins?

Solution

note: we are suppose to answer one question at a time. so i will answer the first. for next repost the question again

To get the 4th head on the 7th toss it means that EXACTLY 3 heads must occur in the first 6 tosses and the 7th toss must be a head

To find the probability of exactly three heads from 6 tosses
Use the binomial formula P(exactly k successes from n events)
= nCk * p(k)^k * (1-p(k))^(n-k)

Where n = the number of events = 6 tosses
k = the number of successes = 3 heads
p(k) = the probability of getting a head = 1/2
1-p(k) = the probability of a failure (tail) = 1/2
n- k = the number of tails = 6 - 3 = 3

Therefore p(exactly 3 heads from 6 tosses) = 6C3 * (1/2)^3 * (1/2)^3
= 6! / (3!*3!) * (1/2)^3 * (1/2)^3
= (6 * 5 * 4 * 3!) / (3! * 3 * 2 * 1) * (1/8) * (1/8)
= 120 / 6 * (1/8) * (1/8)
= 20 * (1/8) * (1/8)
= 20/64
= 5/16

So 3 heads out of 6 tosses = 5/16
The probability of the 7th toss also being a head = 1/2

Therefore P(4th head occurs on 7th toss) = 5/16 * 1/2 = 5/32

Answer = 5/32