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JulijaS [17]
3 years ago
13

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?
Mathematics
1 answer:
aleksandrvk [35]3 years ago
3 0

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}

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