The probability that you will win a game is 0.45.
(a) If you play the game 80 times, what is the most likely number of wins?
(b) What are the mean and variance of a binomial distribution with p = 0.45 and N = 80?
Solution
Here $X$ denote the number of wins out of $N=80$ games.
The probability that you will win a game is $p=0.45$.
The probability mass function of $X$ is
$$ \begin{aligned} P(X=x) &= \binom{80}{x} (0.45)^x (1-0.45)^{80-x},\\ &\qquad x=0,1,\cdots, 80. \end{aligned} $$
(a) The most likely number of wins is the value of $X$ at which the probability mass function $P(X=x)$ becomes maximum. The most likely value of $X$ is called the mode.
For Binomial distribution $B(N,p)$, if $m=(N+1)*p$ is an integer then there are two mode at $(m-1)$ and $m$, but if $m=(N+1)*p$ is not an integer then there is a single mode at $\lfloor m\rfloor$.
We have $(N+1)*p = (80+1)*0.45 = 36.45$, which is not an integer. Hence there is a single mode and it is at $\lfloor m\rfloor =\lfloor 36.45\rfloor =36$.
Thus the most likely number of wins is $36$.
(b) Mean and Variance
Mean of Binomial distribution $B(N,p)$ is $N*p$ and variance of Binomial distribution is $N*p*(1-p)$.
The mean of $X$ is
$$ \begin{aligned} \text{mean } &= N*p\\ &= 80*0.45\\ &= 36 \end{aligned} $$
The variance of $X$ is
$$ \begin{aligned} \text{Variance } &= N*p*(1-p)\\ &= 80*0.45*(1- 0.45)\\ &= 19.8 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
