If three dices are thrown then the probability that the sum of the numbers on their uppermost faces to be atleast 5 is

Options

A. `(1)/(53)`
B. `(53)/(54)`
C. `(1)/(54)`
D. `(52)/(53)`

Explanation

To find the probability that the sum of the numbers on the uppermost faces of three dice is at least 5, we can follow these steps:

### Step 1: Determine the total number of outcomes when throwing three dice.
Each die has 6 faces, so when throwing three dice, the total number of outcomes is:
\[ 6 \times 6 \times 6 = 216 \]

### Step 2: Identify the complementary event.
To find the probability of the sum being at least 5, it is easier to first calculate the probability of the complementary event, which is the sum being less than 5. The possible sums that are less than 5 are 3 and 4.

### Step 3: Calculate the number of outcomes for sums less than 5.
- **Sum = 3**: The only combination is (1, 1, 1). This is 1 outcome.
- **Sum = 4**: The combinations are:
- (1, 1, 2) and its permutations: (1, 2, 1), (2, 1, 1). This gives us 3 outcomes.

So, the total number of outcomes for sums less than 5 is:
\[ 1 \text{ (for sum = 3)} + 3 \text{ (for sum = 4)} = 4 \]

### Step 4: Calculate the probability of the complementary event.
The probability that the sum of the numbers on the uppermost faces is less than 5 is:
\[ P(\text{sum} < 5) = \frac{\text{Number of outcomes for sum < 5}}{\text{Total outcomes}} = \frac{4}{216} = \frac{1}{54} \]

### Step 5: Calculate the probability of the desired event.
Now, we can find the probability that the sum is at least 5:
\[ P(\text{sum} \geq 5) = 1 - P(\text{sum} < 5) = 1 - \frac{1}{54} = \frac{54 - 1}{54} = \frac{53}{54} \]

### Final Answer:
The probability that the sum of the numbers on the uppermost faces of three dice is at least 5 is:
\[ \frac{53}{54} \]

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