Answer :
To determine the missing probabilities, we must use the information provided and the expected value calculation.
1. Let [tex]\( p_{25} \)[/tex] be the probability of scoring 25 points, and [tex]\( p_{50} \)[/tex] the probability of scoring 50 points.
2. According to the probability rule, the sum of the probabilities for all scoring regions must equal 1:
[tex]\[ 0.65 + p_{25} + p_{50} = 1 \][/tex]
Simplifying this equation, we get:
[tex]\[ p_{25} + p_{50} = 0.35 \][/tex]
3. The expected number of points scored per game is given as 16.5. The expected value formula for discrete random variables is:
[tex]\[ E(X) = (0.65 \times 10) + (p_{25} \times 25) + (p_{50} \times 50) = 16.5 \][/tex]
4. Substituting 0.65 for the probability of scoring 10 points, we have:
[tex]\[ 6.5 + 25p_{25} + 50p_{50} = 16.5 \][/tex]
5. Isolating the terms involving [tex]\( p_{25} \)[/tex] and [tex]\( p_{50} \)[/tex]:
[tex]\[ 25p_{25} + 50p_{50} = 10 \][/tex]
6. From our earlier equation, we know [tex]\( p_{25} + p_{50} = 0.35 \)[/tex]. We solve for [tex]\( p_{25} \)[/tex]:
[tex]\[ p_{25} = 0.35 - p_{50} \][/tex]
7. Substitute [tex]\( p_{25} \)[/tex] in the expected value equation:
[tex]\[ 25(0.35 - p_{50}) + 50p_{50} = 10 \][/tex]
Simplifying:
[tex]\[ 8.75 - 25p_{50} + 50p_{50} = 10 \][/tex]
[tex]\[ 8.75 + 25p_{50} = 10 \][/tex]
[tex]\[ 25p_{50} = 1.25 \][/tex]
[tex]\[ p_{50} = \frac{1.25}{25} = 0.05 \][/tex]
8. With [tex]\( p_{50} \)[/tex] found, substitute it back to find [tex]\( p_{25} \)[/tex]:
[tex]\[ p_{25} = 0.35 - 0.05 = 0.30 \][/tex]
So, the probabilities are:
[tex]\[ \begin{aligned} & p_{25} = 0.3 \\ & p_{50} = 0.05 \end{aligned} \][/tex]
Thus, the completed table of probabilities would be:
\begin{tabular}{|l|l|}
\hline
Score & Probability \\
\hline
10 & 0.65 \\
\hline
25 & 0.30 \\
\hline
50 & 0.05 \\
\hline
\end{tabular}
1. Let [tex]\( p_{25} \)[/tex] be the probability of scoring 25 points, and [tex]\( p_{50} \)[/tex] the probability of scoring 50 points.
2. According to the probability rule, the sum of the probabilities for all scoring regions must equal 1:
[tex]\[ 0.65 + p_{25} + p_{50} = 1 \][/tex]
Simplifying this equation, we get:
[tex]\[ p_{25} + p_{50} = 0.35 \][/tex]
3. The expected number of points scored per game is given as 16.5. The expected value formula for discrete random variables is:
[tex]\[ E(X) = (0.65 \times 10) + (p_{25} \times 25) + (p_{50} \times 50) = 16.5 \][/tex]
4. Substituting 0.65 for the probability of scoring 10 points, we have:
[tex]\[ 6.5 + 25p_{25} + 50p_{50} = 16.5 \][/tex]
5. Isolating the terms involving [tex]\( p_{25} \)[/tex] and [tex]\( p_{50} \)[/tex]:
[tex]\[ 25p_{25} + 50p_{50} = 10 \][/tex]
6. From our earlier equation, we know [tex]\( p_{25} + p_{50} = 0.35 \)[/tex]. We solve for [tex]\( p_{25} \)[/tex]:
[tex]\[ p_{25} = 0.35 - p_{50} \][/tex]
7. Substitute [tex]\( p_{25} \)[/tex] in the expected value equation:
[tex]\[ 25(0.35 - p_{50}) + 50p_{50} = 10 \][/tex]
Simplifying:
[tex]\[ 8.75 - 25p_{50} + 50p_{50} = 10 \][/tex]
[tex]\[ 8.75 + 25p_{50} = 10 \][/tex]
[tex]\[ 25p_{50} = 1.25 \][/tex]
[tex]\[ p_{50} = \frac{1.25}{25} = 0.05 \][/tex]
8. With [tex]\( p_{50} \)[/tex] found, substitute it back to find [tex]\( p_{25} \)[/tex]:
[tex]\[ p_{25} = 0.35 - 0.05 = 0.30 \][/tex]
So, the probabilities are:
[tex]\[ \begin{aligned} & p_{25} = 0.3 \\ & p_{50} = 0.05 \end{aligned} \][/tex]
Thus, the completed table of probabilities would be:
\begin{tabular}{|l|l|}
\hline
Score & Probability \\
\hline
10 & 0.65 \\
\hline
25 & 0.30 \\
\hline
50 & 0.05 \\
\hline
\end{tabular}
