If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

[BTGmoderatorDC]

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?


A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6


OA C

Source: Veritas Prep

[swerve]

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?


A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6


OA C

Source: Veritas Prep

Probability if getting \(3\) on a die is: \(P(3)=\dfrac{1}{6}\)
Probability of not getting \(3\) on a die is \(nP(3)=\dfrac{5}{6}\)

Let the three dies be denoted by \(P1, P2,\) and \(P3\)

Probability of getting \(3\) on the first die and not getting \(3\) on the other two dies is given by:

\(=P1(3) \cdot nP(3) \cdot nP(3)\)
\(=\dfrac{1}{6} \cdot \dfrac{5}{6} \cdot \dfrac{5}{6}\)
\(=\dfrac{25}{216}\)

As there are \(3\) dies, so similarly for the rest of the two dies probability will be \(\dfrac{25}{216}\)

Summing up all the \(3\) cases we get:
\(=3\cdot \dfrac{25}{216}\)
\(=\dfrac{25}{72}\)

Hope this help!