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