Question

While staying in a 15-story hotel, Polya plays the following game. She enters an elevator on the 6th floor. She flips a fair coin five times to determine her next five stops. Each time she flips "heads, " she goes up one floor. Each time she flips "tails, " she goes down one floor. What is the probability that each of her next five stops is on the 7th floor or higher? Express your answer as a common fraction.

Answer 1

Answer:

The probability is $\frac{1}{2}$

Step-by-step explanation:

We are going to modeled this problem from the number of heads.

Let be H : ''The number of heads after five flips''

H can assume the following values :

h = 0

h = 1

h = 2

h = 3

h = 4

h = 5

H ~ Bi (n,p)

H ~ Bi (5,0.5)

H can be modeled as a Binomial random variable.

In which p = 0.5 is the success probability (the probability of head in one flip of a fair coin)

n = 5 is the number of times that we flip the coin.

We are also assuming independence in each flip.

The probability function for H is

$P(H=h)=(nCh)p^{h}(1-p)^{n-h}$

$P(H=h)=(5Ch)0.5^{h}0.5^{5-h}$

For all the possible values of H

H ≥ 3 is the event that puts Polya on the 7th floor or higher.

Now we calculate the probability of H ≥ 3

$P(H\geq 3)=P(H=3)+P(H=4)+P(H=5)$

$P(H\geq 3)=(5C3)0.5^{3}0.5^{2}+(5C4)0.5^{4}0.5^{1}+(5C5)0.5^{5}0.5^{0}$

$P(H\geq 3)=0.3125+0.15625+0.03125=0.5$

$P(H\geq 3)=\frac{1}{2}$

Therefore, the probability of each of her next five stops is on the 7th floor or higher is $\frac{1}{2}$