Three dice are rolled 4 times . The probability of getting sum 17 exactly 3 times is

Options

A. `(4 xx 71)/(72^(4))`
B. `(71)/(72^4)`
C. `(71)/(72^4)`
D. `(1)/(72^4)`

Answer

To solve the problem of finding the probability of getting a sum of 17 exactly 3 times when rolling three dice 4 times, we can follow these steps:

### Step 1: Determine the probability of getting a sum of 17 with three dice.
The possible combinations to achieve a sum of 17 with three dice are:
1. (6, 6, 5)
2. (6, 5, 6)
3. (5, 6, 6)

Thus, there are 3 successful outcomes.

The total number of outcomes when rolling three dice is \(6^3 = 216\).

So, the probability \(P\) of getting a sum of 17 is:
\[ P = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{3}{216} = \frac{1}{72} \]

### Step 2: Calculate the probability of not getting a sum of 17.
The probability of not getting a sum of 17, denoted as \(Q\), is:
\[ Q = 1 - P = 1 - \frac{1}{72} = \frac{71}{72} \]

### Step 3: Define the parameters for the binomial probability formula.
We are rolling the dice 4 times, so:
- Number of trials \(n = 4\)
- Number of successes \(r = 3\)
- Probability of success \(p = \frac{1}{72}\)
- Probability of failure \(q = \frac{71}{72}\)

### Step 4: Use the binomial probability formula.
The probability of getting exactly \(r\) successes in \(n\) trials is given by the binomial formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

Substituting the values:
\[ P(X = 3) = \binom{4}{3} \left(\frac{1}{72}\right)^3 \left(\frac{71}{72}\right)^{4-3} \]

### Step 5: Calculate the binomial coefficient and the probabilities.
First, calculate the binomial coefficient:
\[ \binom{4}{3} = 4 \]

Now, substitute into the formula:
\[ P(X = 3) = 4 \left(\frac{1}{72}\right)^3 \left(\frac{71}{72}\right)^1 \]

Calculating \(\left(\frac{1}{72}\right)^3\):
\[ \left(\frac{1}{72}\right)^3 = \frac{1}{373248} \]

Calculating \(\left(\frac{71}{72}\right)^1\):
\[ \left(\frac{71}{72}\right)^1 = \frac{71}{72} \]

Now, combine these results:
\[ P(X = 3) = 4 \cdot \frac{1}{373248} \cdot \frac{71}{72} \]

### Step 6: Final calculation.
\[ P(X = 3) = \frac{4 \cdot 71}{373248 \cdot 72} = \frac{284}{26873856} \]

This fraction can be simplified if necessary, but this is the probability of getting a sum of 17 exactly 3 times when rolling three dice 4 times.