Question

a point is randomly chosen in the square. which statements are true? check all that apply. the point is most likely to be in the shaded region. the point is more likely to be inside the large triangle than outside the large triangle. the probability of the point being in the shaded region is 0.75. the probability that the point is in the square is 1.

Explanation:

Step1: Analyze the square and shaded region

The point is chosen within the square, so the sample space is the square. Let's assume the square has side length \( s \), area \( A_{square}=s^2 \). The shaded region: notice the large triangle and the two smaller triangles. The large triangle (shaded) – wait, actually, the square's area and the shaded area. Wait, the square: the unshaded part is a triangle? Wait, no, looking at the diagram, the shaded region: let's recall that in a square, if we have a triangle from one corner to the midpoints? Wait, no, the key is: the point is in the square, so the probability of being in the square is 1 (since it's chosen there). Now, the shaded region: let's calculate areas. Let the square have side length \( 2 \) (for simplicity, midpoints). Then area of square is \( 2\times2 = 4 \). The unshaded triangle: base \( 2 \), height \( 1 \)? Wait, no, the two smaller triangles: each has base \( 1 \), height \( 2 \)? Wait, maybe better: the large shaded triangle and the two smaller shaded triangles? Wait, actually, the shaded region: let's see, the square is divided. Wait, the correct approach: the probability of being in the square is 1 (since the point is chosen in the square). Now, the shaded region: let's find the area of shaded vs unshaded. Suppose the square has side \( a \). The unshaded triangle: base \( a \), height \( \frac{a}{2} \)? No, looking at the diagram, the vertical side is divided into two equal parts (midpoints). So the unshaded triangle has base \( a \) (top side of square) and height \( \frac{a}{2} \)? Wait, no, the square's height is \( a \), and the unshaded triangle is formed by the top side and the midpoint. Wait, maybe the shaded area: the square area minus the unshaded triangle. Wait, unshaded triangle: base \( a \), height \( \frac{a}{2} \), area \( \frac{1}{2}\times a\times\frac{a}{2}=\frac{a^2}{4} \). So shaded area is \( a^2 - \frac{a^2}{4}=\frac{3a^2}{4} \), so probability of shaded is \( \frac{3}{4}=0.75 \). Now, the large triangle: the big shaded triangle – wait, the large triangle (the biggest shaded one) has area \( \frac{1}{2}\times a\times a=\frac{a^2}{2} \), and the other two shaded triangles: each has area \( \frac{1}{2}\times\frac{a}{2}\times a=\frac{a^2}{4} \)? Wait, no, maybe I messed up. Wait, the square: let's take side length 2. Then square area is 4. The unshaded triangle: base 2, height 1 (since the vertical side is split into two 1s). Area of unshaded: \( \frac{1}{2}\times2\times1 = 1 \). So shaded area is \( 4 - 1 = 3 \), so probability of shaded is \( \frac{3}{4}=0.75 \). Now, the large triangle (the big shaded triangle) has area \( \frac{1}{2}\times2\times2 = 2 \), and the other two shaded triangles: each has area \( \frac{1}{2}\times1\times2 = 1 \), but wait, no, the two smaller triangles: base 1, height 2, so each area 1, but 1+1+2=4? No, square area is 4. Wait, maybe the large triangle is \( \frac{1}{2}\times2\times2 = 2 \), and the two smaller triangles: each is \( \frac{1}{2}\times1\times2 = 1 \), so total shaded 2+1+1=4? No, that can't be. Wait, I think I misread the diagram. Actually, the square: the shaded region is three triangles? Wait, no, the diagram shows a square with a diagonal? No, the lines are from the bottom left corner to the midpoints of the right side. So the right side is split into two equal parts (midpoints). So the square has vertices: let's label them as (0,0), (2,0), (2,2), (0,2). The midpoints of the right side (x=2) are (2,1) and (2,2)? No, (2,1) is the midpoint (since from (2,0) to (2,2), midpoint is (2,1)). Wait, the lines are from (0,0) to (2,1) and from (0,2) to (2,1)? No, the diagram shows a triangle from (0,0) to (2,2) and two triangles from (0,0) to (2,1) and (0,2) to (2,1)? Wait, maybe the correct way: the shaded region is the entire square except a small triangle. Wait, the problem's options:

1. The point is most likely to be in the shaded region: shaded area is 3/4, unshaded is 1/4, so yes, more likely in shaded.

2. The point is more likely to be inside the large triangle than outside: large triangle area – if large triangle is (0,0), (2,2), (0,2), area is \( \frac{1}{2}\times2\times2 = 2 \), outside large triangle in shaded region: shaded area is 3, so outside large triangle in shaded is 3 - 2 = 1. So inside large triangle (2) vs outside (1) in shaded? Wait, no, the total area: square is 4. Inside large triangle: 2, outside large triangle: 4 - 2 = 2. Wait, that's equal. So this statement is false.

3. The probability of the point being in the shaded region is 0.75: as calculated, shaded area 3, square area 4, 3/4=0.75, so true.

4. The probability that the point is in the square is 1: since the point is chosen in the square, this is true (certainty).

So let's check each option:

- "The point is most likely to be in the shaded region": shaded area 3/4, unshaded 1/4, so yes, most likely. True.

- "The point is more likely to be inside the large triangle than outside the large triangle": large triangle area 2, outside large triangle in square: 4 - 2 = 2. So equal, so not more likely. False.

- "The probability of the point being in the shaded region is 0.75": 3/4=0.75, true.

- "The probability that the point is in the square is 1": since the point is chosen in the square, probability is 1. True.

Wait, but let's re-express:

First, the square is the sample space, so \( P(\text{in square}) = 1 \) (fourth option: true).

Shaded area: let's recalculate. Let square have side length \( s \). The unshaded triangle: base \( s \), height \( \frac{s}{2} \) (since the vertical side is split into two equal parts, so height from the top side to the midpoint is \( \frac{s}{2} \)). Area of unshaded triangle: \( \frac{1}{2} \times s \times \frac{s}{2} = \frac{s^2}{4} \). Area of square: \( s^2 \). So shaded area: \( s^2 - \frac{s^2}{4} = \frac{3s^2}{4} \). Thus, \( P(\text{in shaded}) = \frac{3}{4} = 0.75 \) (third option: true).

First option: shaded area is 3/4, unshaded 1/4, so most likely in shaded: true.

Second option: large triangle – let's say the large triangle is the one with base \( s \) and height \( s \) (from (0,0) to (s,s) to (0,s)), area \( \frac{1}{2}s^2 \). Outside this large triangle in the square: area \( s^2 - \frac{1}{2}s^2 = \frac{1}{2}s^2 \). So inside and outside are equal, so not more likely inside. So second option is false.

So the true statements are:

- The point is most likely to be in the shaded region.

- The probability of the point being in the shaded region is 0.75.

- The probability that the point is in the square is 1.

Wait, but let's check the options as given:

Options:

1. The point is most likely to be in the shaded region. – True.

2. The point is more likely to be inside the large triangle than outside the large triangle. – False.

3. The probability of the point being in the shaded region is 0.75. – True.

4. The probability that the point is in the square is 1. – True.

So we need to check all that apply.

Step2: Evaluate each option

- Option 1: Shaded area is \( \frac{3}{4} \), unshaded \( \frac{1}{4} \), so most likely in shaded. True.

- Option 2: Large triangle area \( \frac{1}{2}s^2 \), outside large triangle in square: \( \frac{1}{2}s^2 \). Equal, so not more likely. False.

- Option 3: \( P(\text{shaded}) = \frac{3}{4} = 0.75 \). True.

- Option 4: The point is chosen in the square, so \( P(\text{in square}) = 1 \). True.

Answer:

The true statements are:

• The point is most likely to be in the shaded region.
• The probability of the point being in the shaded region is 0.75.
• The probability that the point is in the square is 1.

(In boxed form for each correct option, but since it's multiple, list them as per the options' text. Assuming the options are labeled as:

A. The point is most likely to be in the shaded region.
B. The point is more likely to be inside the large triangle than outside the large triangle.
C. The probability of the point being in the shaded region is 0.75.
D. The probability that the point is in the square is 1.

Then the correct options are A, C, D.)

\boxed{A, C, D}