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Dafna1 [17]
3 years ago
9

Cole has 4 regular dice and 1 unique dice that has the value 1 on each of the 6 faces. He picks one at random and rolls 3 ones i

n a row. What is the probability he picked the unique dice
Mathematics
1 answer:
marshall27 [118]3 years ago
4 0

Answer:

0.008

Step-by-step explanation:

The probability of having chosen the unique dice three times in a row would be the multiplication of the probability of each event.

The event is always repeated. Take the single dice of 5 dice in total. Therefore the probability is 1/5.

Then the final probability would be:

<em>(1/5) * (1/5) * (1/5)</em>, and that is equal to 0.008.

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\dfrac{ \dfrac{x+3}{4x^2-16} }{ \dfrac{2x^2+10x+12}{2x-4} }

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Write the divide fraction horizontally:
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= \dfrac{x+3}{4x^2-16} \div \dfrac{2x^2+10x+12}{2x-4}

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Factorise the numerators and denominators when possible:
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= \dfrac{x+3}{4(x+2)(x-2)} \div \dfrac{ 2(x + 3) (x + 2)}{2(x-2)}

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Convert the divide fraction to multiplication fraction
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= \dfrac{x+3}{4(x+2)(x-2)} \times \dfrac{2(x-2)}{2(x+3)(x+2)}

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Cancel the factors
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= \dfrac{1}{4(x+2)} \times \dfrac{1}{(x+2)}

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Combine to single fraction
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= \dfrac{1}{4(x+2)^2}



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