Flips of a Coin

P(E) = 1 - P(1 flip or 2 flips or 3 flips or 4 flips)

    = 1 - 1/2+1/4+1/8+1/16
    = 1 - 15/16
    = 1/16

a+b = 17

we have $\text{da answer}=\sum_{i=5}^\infty\frac1{2^i}=\frac1{16}.$

The event that it will take 5+ flips to get a heads is equivalent to the odds of getting four tails in a row. The chances of this are one in 2 to the power of 4, or 1 in 16. Adding the numerator and denominator results in 17.

For the coin to take $5$ heads or more to get heads, we should get $4$ tails in the first four tosses. The probability for this to occur is $\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{2^4} = \dfrac{1}{16}$ giving us the answer $1 + 16 = \boxed{17}$

The events described in the problem will occur if and only if the first four flips are tails. The probability of this is clearly $\left(\dfrac12\right)^4 = \dfrac{1}{16},$ so the answer is $1+16 = \boxed{17}.$

If we have $5$ or more flips before we get heads, then the first $4$ flips must be tails. The probability of getting a tail on any given flip is $\frac{1}{2}$ , so the probability of $4$ tails in a row is $\left(\frac{1}{2}\right)^4$ , which is equal to $\frac{1}{16}$ . $1+16=17$ .

1) there can be infinite possibilities as shown below. (T = tails and H = heads) T T T T H , T T T T T H , T T T T T T H , T T T T T T T H, ............

2)for 1st case probability is 1/32....for second case it is 1/64....for third case its 1/128 and so on.

3) since any of these is possible (because its given in question 5 or more heads) we wilh to add these as follows.

(1/32) + (1/64) + (1/128) + ....... this forms a GP with the first term (1/32) and common difference (1/2)

4) find the sum to infinite terms we get (1/16)....therefore a+b = 17

Needing 5 or more tosses to get heads is equivalent to the first four tosses giving tails. The chance of this is 1/2^4=1/16. 1+16=17

We have to read the question carefully. It is 5 or more flips. What is means is that 5,6,7,8,.... infinite number of flips that does not give u a heads. For 5 flips, it is written as (1/2)^5 and we have to consider 6,7,8..... which is just (1/2)^6, (1/2)^7, (1/2)^8..... and since it is a geometric progression of 1/2 starting from n = 5 to infinity, (1/32)/(1-1/2) = 1/16