Question

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?

Answer 1

`(1)/(4)`

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
`\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 \}`

Let E denotes the event: 6 occurs

Let F denotes the event that 6 does not occur .

Given: P(E) =
`(1)/(4)`

So, P(F) =
`1-(1)/(4)=(3)/(4)`

Therefore, probability of rolling at least two sixes =
`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 )`

`=\left ( (1)/(4) \right )\left ( (1)/(4) \right )\left ( (3)/(4) \right )+\left ( (1)/(4) \right )\left ( (1)/(4) \right )\left ( (3)/(4) \right )+\left ( (1)/(4) \right )\left ( (1)/(4) \right )\left ( (3)/(4) \right )+\left ( (1)/(4) \right )\left ( (1)/(4) \right )\left ( (3)/(4) \right )+\left ( (1)/(4) \right )\left ( (1)/(4) \right )\left ( (3)/(4) \right )+\left ( (1)/(4) \right )\left ( (1)/(4) \right )\left ( (1)/(4) \right )`

`=(3)/(64)+(3)/(64)+(3)/(64)+(3)/(64)+(3)/(64)+(1)/(64)\\\\=(16)/(64)\\\\=(1)/(4)`