Chapter2.Rmd

---
title: "Chapter 2: Discrete random variables"
output: html_notebook
---


# Binomial Distribution 

* A binomial experiment consists of a fixed number of $n$ trials. 
* Each trial has two possible outcomes: success, failure. 
* The probability of success is denoted by $p,\:0\leq p\leq 1$.
* The probability of failure is $1-p$. 
* The trials are assumed independent:
* the outcome of one trial does not affect the  outcome probability of any other trial.


```{r}
n=10  # Sample size
p=.5 # probability of success 
x=0:n  # Possible values
round(dbinom(x,n,p),4)

choose(10,6)

210*(0.5)^4*(0.5)^6

.8^10
dbinom(0,10,0.2)

45*(.2)^2*(.8)^8
choose(10,2)*(.2)^2*(.8)^8
dbinom(2,10,.2)
sum(dbinom(0:4,10,.2))
pbinom(4,10,.2)
library(formattable)
dbinom(2,10,.2)
percent(dbinom(2,10,.2))






```


```{r}
B<-as.table(dbinom(0:10,10,.2))
B
names(B)<-0:10
B
plot(B,col="red",main="Bar plot for binomial PDF",xlab="Number of correct answers",ylab="Probability")
```
```{r}
set.seed(123)
res<-rbinom(10000,10,.2)
head(res,10)
mean(res)
round(dbinom(0:10,10,.2),4)
table(res)
sum((0:10)*dbinom(0:10,10,.2))
(0:10)%*%dbinom(0:10,10,.2)

var(res)

mean((res-mean(res))^2)
(0:10-2)^2%*%dbinom(0:10,10,.2)
10*.2*.8
```




## Exercice 7.84
 
On a $n=10$ er $p=0.3$.

a. Calculer $\Pr(X=3)$.
```{r}
round(dbinom(3,10,0.3),4)
```

b. Calculer $\Pr(X=5)$.

```{r}
round(dbinom(5,10,0.3),4)
```


c. Calculer $\Pr(X=8)$.

```{r}
round(dbinom(8,10,0.3),4)
```

## Obtenir les 3 résulats d'un coup.


```{r}
round(dbinom(c(3,5,8),10,0.3),4)
```

## Executer le Chunk en ligne de manière transparente

a. Calculer $\Pr(X=3)$. On trouve $\Pr(X=3)=$ `r round(dbinom(3,10,0.3),4)`.

## Exercice 7.92



```{r}
dbinom(1,4,1/4)
dbinom(2,8,1/4)
dbinom(3,12,1/4)

```


Taper un code en ligne. Un exemple simple `r 112+234`.



## 7.101 b.

$$E(X)=100\times \frac{244}{495}$$


On trouve `r round(100*244/495,0)`. Il gagne environ en moyenne `r round(100*244/495,0)` parties pour 100 parties jouées.



## Exercice 7.138


Soit $X$ le nombre de faux-positifs pour 10 ans de mammographie. On a $X$ suit une loi $\mathcal{B}(10,p)$ où $p$ est inconnu. On sait que $\Pr(X\geq 1)=0.6$. On doit trouver $p$. 



On a $$Pr(X \geq 1) = 1 - Pr(X=0)$$ Soit :

$$
1-(1-p)^{10}=0.60.
$$
Cela revient à résoudre l'équation suivante : 
$$
(1-p)^{10}=0.4,
$$
qui donne 
$$
1-p = 0.4^{1/10}
$$.
Ainsi, 
$$p = 1 - 0.4^{1/10}
$$
Chaque année la probabilité qu'une femme qui passe une mamographie obtienne un faux positif s'élève à `r percent(1 - 0.4^(1/10),1)`.