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

`P(T\ n\ S) = (1)/(14)`

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) = (n(T))/(Total)`

`P(T) = (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) = (n(S))/(Total)`

`P(S) = (3)/(14)`

Recall that

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

`P(T\ n\ S) = (5)/(15) * (3)/(14)`

`P(T\ n\ S) = (15)/(15 * 14)`

`P(T\ n\ S) = (1)/(14)`

Hence, the required probability is

`P(T\ n\ S) = (1)/(14)`