Answer:
[tex]P(T_1\ n\ T_2) = \frac{2}{21}[/tex]
Step-by-step explanation:
From the comments in your question; we have
Stars = 4
Triangles = 5
Circles = 3
Squares = 3
Required
Determine the probability of both shapes being triangles
First, calculate the total
[tex]Total = 4 + 5 + 3 + 3[/tex]
[tex]Total = 15[/tex]
Next, calculate the probability of the first selected shape being a triangle;
P(T₁) = Number of triangles divided by total number of shapes
[tex]P(T_1) = \frac{5}{15}[/tex]
[tex]P(T_1) = \frac{1}{3}[/tex]
Next, calculate the probability of the second selected shape being a triangle;
P(T₂) = Number of triangles divided by total number of shapes
Because it's probability without replacement. the number of triangle left is 4 and the number of shapes left is 14;
So:
[tex]P(T_2) = \frac{4}{14}[/tex]
[tex]P(T_2) = \frac{2}{7}[/tex]
Hence:
[tex]P(T_1\ n\ T_2) = P(T_1) * P(T_2)[/tex]
[tex]P(T_1\ n\ T_2) = \frac{1}{3} * \frac{2}{7}[/tex]
[tex]P(T_1\ n\ T_2) = \frac{2}{21}[/tex]
Hence, the required probability is [tex]\frac{2}{21}[/tex]
Answer:
2/21
Step-by-step explanation:
There are 4 stars, 5 triangles, 3 circles, and 3 squares
For the first drawing there are 5 triangles and a total of 15 shapes.
p(first triangle) = 5/15 = 1/3
Since there is no replacement, for the second drawing, and one triangle has been taken, now there are 4 triangles left and 14 total shapes left.
p(second triangle) = 4/14 = 2/7
p(two triangles) = p(first triangle) * p(second triangle)
p(two triangles) = 1/3 * 2/7
p(two triangles) = 2/21
