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Now, the applet can be used to simulate ten times how the points will be distributed on the segment. The results of each trial will be recorded in a table. To do so, let $N_L$ be the number of points to the left and $N_R$ be the number of points to the right of $(a,2).$ Here, an example simulation is shown.

Trial	$N_L$	$N_R$
$1$	$8$	$2$
$2$	$7$	$3$
$3$	$9$	$1$
$4$	$7$	$3$
$5$	$6$	$4$
$6$	$6$	$4$
$7$	$7$	$3$
$8$	$9$	$1$
$9$	$6$	$4$
$10$	$7$	$3$

On average, there are $7.2$ points to the left of $(a,2)$ on the horizontal segment. In other words, the probability of a random point appearing on the left side is about the ratio of $7.2$ to $10.$ The probability of a point between $0$ and $a$ on the horizontal segment can also be found using the geometric probability. It is equal to the ratio of the length of the segment between $0$ and $a$ to the length of the horizontal segment. The ratio obtained using the simulation data $\frac{7.2}{10}$ is an estimate of the probability obtained using geometric measures $\frac{a}{1}.$ With this in mind, the value of $a$ can be calculated. Therefore, $a,$ the value of $\sqrt{2},$ is about $1.44.$ Notice that this is close to the value of $\sqrt{2}.$

The results can be recorded in a table. Here is an example table.

Trial	$N_U$	$N_A$
$1$	$35$	$65$
$2$	$39$	$61$
$3$	$34$	$66$
$4$	$33$	$67$
$5$	$31$	$69$
$6$	$28$	$72$
$7$	$27$	$73$
$8$	$37$	$63$
$9$	$33$	$67$
$10$	$35$	$65$

On average, there are $33.2$ points under the curve. In other words, the probability of a random point appearing under the curve is about the ratio of $33.2$ to $100.$ The probability of a point being under the curve can also be found using the geometric probability. It is equal to the ratio of the area under the curve $A_U$ to the area of the unit square $1.$ The ratio obtained using the simulation data $\frac{33.2}{100}$ is an estimate of the probability obtained using geometric measures $\frac{A_U}{1}.$ With this in mind, the value of $A_U$ can be calculated. Therefore, the area under the graph of $y=x^2$ from $0$ to $1$ is about $0.332$ square units. This result is close to the exact value of the area.

Recall the rules of the game. For the given table, the number of coins Vincenzo and his friends can get is found as follows. Since paying $2$ coins is a loss, its effect will be negative. When the results are as shown in the table, they will get $27$ coins.

When Jordan does not know the correct answer and guesses, she either receives $1$ point or loses $0.5$ points. Let $X$ be the random variable that represents the point value assigned to answers. Then, $X$ can be written as follows. Since there are $5$ choices for each question, the probability of selecting the right answer is $\frac{1}{5}$ and the probability of selecting an incorrect one is $\frac{4}{5}.$ The following table shows probabilities.

$X$	$1$	$\text{-}0.5$
$P(X)$	$\frac{1}{5}$	$\frac{4}{5}$

From here, the expected value of the random variable $X$ can be calculated. Because the expected value for each guess is $\text{-}0.2,$ Jordan would lose $0.2$ points for each guess. Therefore, she should not guess.

Note that there is only one way to select $4$ females out of the seven. Since the order of selecting people does not matter, the number of possible outcomes can be found using combinations. The number of combinations when selecting $r$ items out of $n$ is given by the Combination Formula. By substituting $7$ for $n$ and $4$ for $r,$ the number of combinations can be calculated. There are $35$ ways of selecting $4$ people out of $7.$ This is the number of possible outcomes. Therefore, the probability of selecting $4$ females is given by the ratio of $1$ to $35.$ To find the expected value of $X,$ the probabilities of each case needs to be determined. To do so, combinations and the Fundamental Counting Principle will be used. Now, consider the case in which $1$ female and $3$ males are selected. Here, ${}_4{C}_1$ represents the number of possible ways $1$ female is selected out of $4$ females and ${}_3{C}_3$ represents the number of possible ways $3$ males are selected out of $3$ males. Compute the value of the product. Therefore, there are $4$ possible ways to select $1$ female and $3$ males. Similarly, the rest of the possible cases can be found.

Number of Possible Outcomes, $35$
$1$ Female and $3$ Males	$2$ Females and $2$ Males	$3$ Females and $1$ Male	$4$ Females and $0$ Males
${}_4{C}_1\cdot {}_3{C}_3$	${}_4{C}_2\cdot {}_3{C}_2$	${}_4{C}_3\cdot {}_3{C}_1$	${}_4{C}_4\cdot {}_3{C}_0$
$4\cdot 1$	$6\cdot 3$	$4\cdot 3$	$1\cdot 1$
$4$	$18$	$12$	$1$

Considering the above table, the probabilities can be determined.

$X$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{4}{35}$	$\frac{18}{35}$	$\frac{12}{35}$	$\frac{1}{35}$

From here, the expected value of the random variable $X$ can be computed. This equation means that when $4$ people are randomly selected from a group of $3$ males and $4$ females, approximately $2$ are expected to be female. Therefore, selecting only four females, in this case, is unlikely.