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Two pentagonal solids with faces numbered 3-7 and 6-10 are rolled. What is the probability of getting a sum greater than or equal to 11?

Mathematics

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2 parts
b

2225\frac{22}{25}

12

11 cm11 \text{ cm}

Step-by-step solution

10 steps

Step 1: Identify the numbers on each pentagonal solid.
The first pentagonal solid has sides marked P1={3,4,5,6,7}P_1 = \{3, 4, 5, 6, 7\}.
The second pentagonal solid has sides marked P2={6,7,8,9,10}P_2 = \{6, 7, 8, 9, 10\}.

Part a· 1 step

Draw a probability space to show the possible sums.

  1. Create a table to show all possible sums when the two solids are rolled.

    The table below shows the sum of the numbers appearing on the bottom face.

    P1∖P2678910391011121341011121314511121314156121314151671314151617\begin{array}{|c|c|c|c|c|c|} \hline P_1 \setminus P_2 & 6 & 7 & 8 & 9 & 10 \\ \hline 3 & 9 & 10 & 11 & 12 & 13 \\ \hline 4 & 10 & 11 & 12 & 13 & 14 \\ \hline 5 & 11 & 12 & 13 & 14 & 15 \\ \hline 6 & 12 & 13 & 14 & 15 & 16 \\ \hline 7 & 13 & 14 & 15 & 16 & 17 \\ \hline \end{array}

Part b

2225\frac{22}{25}

Find the probability of getting a sum greater than or equal to 11.

  1. Count the total number of possible outcomes.

    There are 5 outcomes for P1P_1 and 5 outcomes for P2P_2.
    Total number of outcomes = 5×5=255 \times 5 = 25.

  2. Count the number of outcomes where the sum is greater than or equal to 11.

    From the table:

    • For P1=3P_1 = 3: Sums are 11,12,1311, 12, 13 (3 outcomes)
    • For P1=4P_1 = 4: Sums are 11,12,13,1411, 12, 13, 14 (4 outcomes)
    • For P1=5P_1 = 5: Sums are 11,12,13,14,1511, 12, 13, 14, 15 (5 outcomes)
    • For P1=6P_1 = 6: Sums are 12,13,14,15,1612, 13, 14, 15, 16 (5 outcomes)
    • For P1=7P_1 = 7: Sums are 13,14,15,16,1713, 14, 15, 16, 17 (5 outcomes) Number of favorable outcomes = 3+4+5+5+5=223 + 4 + 5 + 5 + 5 = 22.

  3. Calculate the probability.

    P(sum≥11)=Number of favorable outcomesTotal number of outcomes=2225P(\text{sum} \ge 11) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{22}{25}

    The probability of getting a sum greater than or equal to 11 is 2225\boxed{\frac{22}{25}}.

Part 12

11 cm11 \text{ cm}

  1. Identify the given information.

    Diameter of the soccer ball = 30 cm.
    The ball is placed at a corner of a rectangular room.

  2. Calculate the radius of the ball.

    Radius (r)=Diameter2=30 cm2=15 cm\text{Radius } (r) = \frac{\text{Diameter}}{2} = \frac{30 \text{ cm}}{2} = 15 \text{ cm}

  3. Determine the coordinates of the center of the ball.

    When the ball is placed in a corner, it touches the three mutually perpendicular planes (floor and two walls). The center of the ball will be at a distance equal to its radius from each of these planes.
    If the corner of the room is at the origin $(0,0,0)$, the center of the ball will be at $(r, r, r)$.
    So, the center of the ball is at $(15, 15, 15)$.

  4. Calculate the distance from the corner of the room to the center of the ball.

    Using the distance formula in 3D:

    Dc=(x2−x1)2+(y2−y1)2+(z2−z1)2D_c = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

    Here, (x1,y1,z1)=(0,0,0)(x_1, y_1, z_1) = (0,0,0) (corner) and (x2,y2,z2)=(15,15,15)(x_2, y_2, z_2) = (15,15,15) (center of ball).

    Dc=(15−0)2+(15−0)2+(15−0)2D_c = \sqrt{(15-0)^2 + (15-0)^2 + (15-0)^2} Dc=152+152+152D_c = \sqrt{15^2 + 15^2 + 15^2} Dc=3×152D_c = \sqrt{3 \times 15^2} Dc=153 cmD_c = 15\sqrt{3} \text{ cm}

  5. Calculate the shortest distance from the ball to the corner of the room.

    The shortest distance from the surface of the ball to the corner is the distance from the center of the ball to the corner minus the radius of the ball.

    Dshortest=Dc−rD_{\text{shortest}} = D_c - r Dshortest=153−15D_{\text{shortest}} = 15\sqrt{3} - 15 Dshortest=15(3−1)D_{\text{shortest}} = 15(\sqrt{3} - 1)

    Using 3≈1.73205\sqrt{3} \approx 1.73205:

    Dshortest=15(1.73205−1)D_{\text{shortest}} = 15(1.73205 - 1) Dshortest=15(0.73205)D_{\text{shortest}} = 15(0.73205) Dshortest=10.98075 cmD_{\text{shortest}} = 10.98075 \text{ cm}

  6. Round the answer to 2 significant figures.

    Dshortest≈11 cmD_{\text{shortest}} \approx 11 \text{ cm}

    The shortest distance of the ball from the corner of the room is 11 cm\boxed{11 \text{ cm}}.

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