Question

Imagine Scott stood at zero on a life-sized number line. His friend flipped a coin 6 times. When the coin

came up heads, he moved one unit to the right. When the coin came up tails, he moved one unit to the left.
After each flip of the coin, Scott's friend recorded his position on the number line. Let f(n) represent Scott's
position on the number line after the nth coin flip.

a. How many different outcomes are there for the sequence of 6 coin tosses?

b. Calculate the probability, before the coin flips have begun, that f(6) = 0, f(6)= 1, and f(6) = 6.

c. Make a bar graph showing the frequency of the different outcomes for this random walk.

d. Which number is Scott most likely to land on after the six coin flips? Why?

Answer 1

Answer: a. The sequence of 6 coin tosses can have 2^6 = 64 different outcomes. This is because each coin flip has two possible outcomes (heads or tails), and there are 6 independent coin flips.

b. To calculate the probability of different outcomes for f(6), we need to consider the number of ways each outcome can occur divided by the total number of possible outcomes.

Probability of f(6) = 0:

To end up at 0, Scott needs to have an equal number of heads and tails. This can happen in two ways: HHTTTT or TTHHHH. So, the probability of f(6) = 0 is 2/64 = 1/32.

Probability of f(6) = 1:

To end up at 1, Scott needs to have 4 tails and 2 heads. This can happen in six ways: TTHHHT, TTHHTH, TTHTHH, TTHTHH, THTTHH, or HTTTHH. So, the probability of f(6) = 1 is 6/64 = 3/32.

Probability of f(6) = 6:

To end up at 6, Scott needs to have 6 heads and no tails, which can only happen in one way: HHHHHH. So, the probability of f(6) = 6 is 1/64.

c. Here's a bar graph showing the frequency of different outcomes for this random walk:

Number of Units (f(6))

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

0 | *

1 | ***

2 |

3 |

4 |

5 |

6 | *

In the above bar graph, the asterisks (*) represent the outcomes with non-zero frequency.

d. Scott is most likely to land on f(6) = 1. This is because there are more ways to achieve f(6) = 1 compared to other outcomes. As calculated in part b, there are 6 different ways to end up at f(6) = 1, while there are only 2 ways to end up at f(6) = 0 and only 1 way to end up at f(6) = 6. Therefore, the highest probability is associated with f(6) = 1, making it the most likely outcome.