Question

If two of these shapes are randomly chosen, one after the other without replacement, what is the probability that the first will be a triangle and the second will be a square?

Answer 1

Answer:

$P(T\ n\ S) = \frac{1}{14}$

Step-by-step explanation:

*Missing Part of the Question*

4 stars

5 triangles

3 circles

3 squares

Required

Determine the probability of triangle being first then square being second

$Total = 4 + 5 + 3 + 3$

$Total = 15$

Represent the triangle with T and square with S

So, we're solving for P(T n S)

$P(T\ n\ S) = P(T) * P(S)$

Solving for P(T)

$P(T) = \frac{n(T)}{Total}$

$P(T) = \frac{5}{15}$

Solving for P(S)

The question implies a probability without replacement;

Hence Total has now been reduced by 1

Total = 14

$P(S) = \frac{n(S)}{Total}$

$P(S) = \frac{3}{14}$

Recall that

$P(T\ n\ S) = P(T) * P(S)$

$P(T\ n\ S) = \frac{5}{15} * \frac{3}{14}$

$P(T\ n\ S) = \frac{15}{15 * 14}$

$P(T\ n\ S) = \frac{1}{14}$

Hence, the required probability is

$P(T\ n\ S) = \frac{1}{14}$