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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 be a triangle and the second will be a square?

User Hylowaker
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1 Answer

4 votes

Answer:


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)

User Jacob Birkett
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