Problem Solving

Veritas Prep

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


A. \(\frac{25}{216}\)

B. \(\frac{50}{216}\)

C. \(\frac{25}{72}\)

D. \(\frac{25}{36}\)

E. \(\frac{5}{6}\)

OA C

AAPL wrote:Veritas Prep

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


A. \(\frac{25}{216}\)

B. \(\frac{50}{216}\)

C. \(\frac{25}{72}\)

D. \(\frac{25}{36}\)

E. \(\frac{5}{6}\)

P(exactly n times) = P(one way) * all possible ways.

Let T = 3 and N = not 3.

P(one way):
One way to get exactly one 3:
TNN.
P(T on the 1st roll) = 1/6. (Of the 6 possible rolls, one is 3.)
P(N on the 2nd roll) = 5/6. (Of the 6 possible rolls, five are not 3.)
P(N on the 3nd roll) = 5/6. (Of the 6 possible rolls, five are not 3.)
Since we want all of these events to happen, we MULTIPLY:
1/6 * 5/6 * 5/6 = 25/216.

All possible ways:
TNN is only ONE WAY to exactly one 3.
Now we must account for ALL OF THE WAYS to get exactly one 3.
Any arrangement of the letters TNN represents one way to exactly one 3.
Thus, to account for ALL OF THE WAYS to get exactly one 3, the result above must be multiplied by the number of ways to arrange the letters TNN.
Number of ways to arrange 3 elements = 3!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical N's:
3!/2! = 3.

Multiplying the results above, we get:
P(exactly one 3) = 25/216 * 3 = 25/72.

The correct answer is C.

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