Question

A fair coin is flipped 10 times. Each time it shows heads, Ann gets a point; otherwise Bob gets a point.

(i) What is the most likely final result?

(ii) Which is more likely: that it ends 5 − 5 or that somebody wins 6 − 4?

(iii) If Ann wins the first three rounds, what is the probability that she ends up the winner?

(iv) If Ann wins the first four rounds, what is the probability that Bob never takes the lead?

(v) What is the probability that the lead changes four times?

--> So far all I've gotten is that the final result is most likely going to be a 5-5 tie. No idea how to do the rest.

Answer 1

PART 1

The process shows a binomial distribution, with the following set of the following parameters:

n = 10:

p=0.5;

q=1-p=0.5:

Such a set of parameters is characteristic of a process involving a fair, unbiased coin.

As is the case with binomial distributions, the most likely event corresponds to the the mean of the distribution:

= np

= 10 X 0.5

$=5$

PART 2

Probability that it's a draw:

$^{10}C_{5}(\frac{1}{2})^{10}$

Here, we have chosen five rounds for A to win. This automatically means that B wins the other five, leading to a draw. We have multiplied by a factor of $(\frac{1}{2})^{10}$ ​​​​​​, to account for the probability of A winning five rounds and the probability of B winning five rounds.

Similarly, the probability of A winning six rounds equates to

$^{10}C_{6} (\frac{1}{2})^{10}$

and that of B winning six rounds equates to

$^{10}C_{4} (\frac{1}{2})^{10}$

Therefore, the probability of someone winning 6-4 is

$^{10}C_{4} (\frac{1}{2})^{10}+^{10}C_{6} (\frac{1}{2})^{10}$​

This equates to 0.4101 (approximately), which is more that the probability that it's a draw, i.e. 0.2461 (approximately).

PART 3

To win, a player must win at least six rounds. This is equivalent to the probability of not losing at least six rounds, which is equal to the 1 - probability of losing at most six rounds.

In the question, Ann has won three rounds, and needs to win three more to win the whole set. This probability equals:

ようつというのストラン メキシコ メン

While this can be calculated, we can make use of the conclusions drawn in the first paragraph of the solution of this part to make our job easier.

Using this, we have:

$1-{} ^{7}C_{0} (\frac{1}{2})^{7}-^{7}C_{1} (\frac{1}{2})^{7}-^{7}C_{2} (\frac{1}{2})^{7}$

This equates to ​​​​​​

PART 4

We have 6 rounds left. Of these, B can win up to four with A always being in the lead. If B wins five games, A must have won a round before, i.e., we can't have A winning a round after B has won all five before. This total probability equates to

$(^{6}C_{0}+^{6}C_{1}+^{6}C_{2}+^{6}C_{3}+^{6}C_{4}+^{6}C_{5}-1) (\frac{1}{2})^6$

where we have subtracted 1 to account for the case where B wins the first 5 of the six rounds placing him in the lead temporarily. The aforementioned value equates to 0.96875.