A weighted die, numbered one through six, has a probability of 1/4 of rolling a six. if this die is rolled three times, and each roll is independent, what is the probability of rolling at least two sixes?

Probability of rolling no sixes = (3/4)^3 = 27/64

Probability of first one being 6 and the rest Not 6 = 1/4 * 3/4* 3/4 = 9/64

THere are 3 ways for this to happen so total probability of 1 six = 27/64

Therefore Probability of 2 or more sixes = 1 - 2(27/64) = 10/54 = 5/27

Answer is 5/27

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berno berno · Answer 2 · 2019-10-22T21:35:46+02:00

Answer:

[tex]\frac{1}{4}[/tex]

Step-by-step explanation:

Three events A, B and C are said to be independent if occurrence of one outcome does not affect the other .

In such case P ( A and B and C ) = P(A) × P(B) × P (C)

Given : A weighted die, numbered one through six, has a probability of 1/4 of rolling a six .

As the die is rolled three times, possible outcomes are [tex]\left \{ \left ( 6,6,1 \right )\,,\,\left ( 6,6,2 \right )\,,\,\left ( 6,6,3 \right )\,,\,\left ( 6,6,4 \right )\,,\,\left ( 6,6,5 \right )\,,\,\left ( 6,6,6 \right ) \right \}[/tex]

Let E denotes the event: 6 occurs

Let F denotes the event that 6 does not occur .

Given: P(E) = [tex]\frac{1}{4}[/tex]

So, P(F) = [tex]1-\frac{1}{4}=\frac{3}{4}[/tex]

Therefore, probability of rolling at least two sixes = [tex]P\left ( 6,6,1 \right )+P\left ( 6,6,2 \right )+P\left ( 6,6,3 \right )+P\left ( 6,6,4 \right )+P\left ( 6,6,5 \right )+P\left ( 6,6,6 \right )[/tex]

[tex]=\left ( \frac{1}{4} \right )\left ( \frac{1}{4} \right )\left ( \frac{3}{4} \right )+\left ( \frac{1}{4} \right )\left ( \frac{1}{4} \right )\left ( \frac{3}{4} \right )+\left ( \frac{1}{4} \right )\left ( \frac{1}{4} \right )\left ( \frac{3}{4} \right )+\left ( \frac{1}{4} \right )\left ( \frac{1}{4} \right )\left ( \frac{3}{4} \right )+\left ( \frac{1}{4} \right )\left ( \frac{1}{4} \right )\left ( \frac{3}{4} \right )+\left ( \frac{1}{4} \right )\left ( \frac{1}{4} \right )\left ( \frac{1}{4} \right )[/tex]

[tex]=\frac{3}{64}+\frac{3}{64}+\frac{3}{64}+\frac{3}{64}+\frac{3}{64}+\frac{1}{64}\\\\=\frac{16}{64}\\\\=\frac{1}{4}[/tex]