The bag contains 4 balls: 1 ball with the letter A and 3 balls with the letter B.
Urn A contains 5 tickets: 3 tickets of 50 euros and 2 tickets of 10 euros.
Urn B contains 4 tickets: 1 ticket of 50 euros and 3 tickets of 10 euros.
A player randomly draws a ball from the bag:
if it is a ball with the letter A, he randomly draws a ticket from urn A.
if it is a ball with the letter B, he randomly draws a ticket from urn B.
We note the following events:
$A$: the player obtains a ball with the letter A.
$C$: the player obtains a 50 euro ticket.
Copy and complete the tree opposite representing the situation.
What is the probability of the event ``the player obtains a ball with the letter A and a ticket of $50 €$''?
Prove that the probability $P(C)$ is equal to 0.3375.
The player obtained a 10 euro ticket. Is the statement ``There is more than $80\%$ chance that he previously obtained a ball with the letter B'' true? Justify.
We denote $X_1$ the random variable that gives the sum, in euros, obtained by the player. Example: if the player obtains a 50 euro ticket, then $X_1 = 50$. Show that the expectation $E(X_1)$ is equal to 23.50 and that the variance $V(X_1)$ is equal to 357.75.
After returning the ball to the bag and the ticket to the urn from which it was taken, the player plays a second game. We denote $X_2$ the random variable that gives the sum obtained by the player in this second game. We denote $Y$ the random variable defined as follows: $Y = X_1 + X_2$. a. Show that $E(Y) = 47$. b. Explain why we have $V(Y) = V(X_1) + V(X_2)$.
The player plays likewise a third, fourth, \ldots, hundredth game. We thus define in the same way the random variables $X_3, X_4, \ldots, X_{100}$. We denote $Z$ the random variable defined by $Z = X_1 + X_2 + \ldots + X_{100}$. Prove that the probability that $Z$ belongs to the interval $]1950; 2750[$ is greater than or equal to 0.75.
We have a bag and two urns A and B.
\begin{itemize}
\item The bag contains 4 balls: 1 ball with the letter A and 3 balls with the letter B.
\item Urn A contains 5 tickets: 3 tickets of 50 euros and 2 tickets of 10 euros.
\item Urn B contains 4 tickets: 1 ticket of 50 euros and 3 tickets of 10 euros.
\end{itemize}
A player randomly draws a ball from the bag:
\begin{itemize}
\item if it is a ball with the letter A, he randomly draws a ticket from urn A.
\item if it is a ball with the letter B, he randomly draws a ticket from urn B.
\end{itemize}
We note the following events:
\begin{itemize}
\item $A$: the player obtains a ball with the letter A.
\item $C$: the player obtains a 50 euro ticket.
\end{itemize}
\begin{enumerate}
\item Copy and complete the tree opposite representing the situation.
\item What is the probability of the event ``the player obtains a ball with the letter A and a ticket of $50 €$''?
\item Prove that the probability $P(C)$ is equal to 0.3375.
\item The player obtained a 10 euro ticket. Is the statement ``There is more than $80\%$ chance that he previously obtained a ball with the letter B'' true? Justify.
\item We denote $X_1$ the random variable that gives the sum, in euros, obtained by the player. Example: if the player obtains a 50 euro ticket, then $X_1 = 50$. Show that the expectation $E(X_1)$ is equal to 23.50 and that the variance $V(X_1)$ is equal to 357.75.
\item After returning the ball to the bag and the ticket to the urn from which it was taken, the player plays a second game. We denote $X_2$ the random variable that gives the sum obtained by the player in this second game. We denote $Y$ the random variable defined as follows: $Y = X_1 + X_2$.\\
a. Show that $E(Y) = 47$.\\
b. Explain why we have $V(Y) = V(X_1) + V(X_2)$.
\item The player plays likewise a third, fourth, \ldots, hundredth game. We thus define in the same way the random variables $X_3, X_4, \ldots, X_{100}$. We denote $Z$ the random variable defined by $Z = X_1 + X_2 + \ldots + X_{100}$. Prove that the probability that $Z$ belongs to the interval $]1950; 2750[$ is greater than or equal to 0.75.
\end{enumerate}