Question

A fair die is rolled four times. What is the probability of obtaining at least three 2's?

Answer 1

The given problem can be modeled as a Binomial distribution since the following conditions are satisfied.

There are only two possible outcomes either you get a 2 or you dont.

The number of trials is fixed. (n = 4)

The probability of success is fixed.

The probability of success is the probability of getting a 2 that is 1/6

The Binomial distribution is given by

`P(x)=^nC_x\cdot p^x\cdot(1-p)^(n-x)`

Where n is the number of trials that is 4

x is the outcome of interest which means getting a 2.

At least three 2's means three or more than three.

So, x = 3, 4

nCx is the number of possible combinations

So, this means that we have to find

`P(x\ge3)=P(x=3)+P(x=4)_{}`

P(x = 3):

`\begin{gathered} P(x=3)=^4C_3\cdot((1)/(6))^3\cdot(1-(1)/(6))^(4-3) \\ P(x=3)=4\cdot((1)/(6))^3\cdot((5)/(6))^1 \\ P(x=3)=0.0154 \end{gathered}`

P(x = 4):

`\begin{gathered} P(x=4)=^4C_4\cdot((1)/(6))^4\cdot(1-(1)/(6))^(4-4) \\ P(x=4)=1\cdot((1)/(6))^4\cdot((5)/(6))^0 \\ P(x=4)=0.00077 \end{gathered}`

So, the probability of obtaining at least three 2's is

`\begin{gathered} P(x\ge3)=P(x=3)+P(x=4)_{} \\ P(x\ge3)=0.0154+0.00077 \\ P(x\ge3)=0.01617 \end{gathered}`

Therefore, the probability of obtaining at least three 2's is 0.01617