3 years ago

A fair charges an admission fee of dollars for each person.

Mathematics

1 answer:

zhuklara [117]3 years ago

6 0

Answer:

mag aral hindi nakaasa sa brainly

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If two of these shapes are randomly chosen, one after the other without replacement, what is the probability that the first will

aleksandrvk [35]

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)

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

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The question implies a probability without replacement;

Hence Total has now been reduced by 1

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

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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

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