A fair coin is tossed once. If it shows head, then 2 fiar disc are thrown simultaneously otherwise 3 fair disc are thrown simultaneously. The probability that all the dice show different numbers is k, then 180 k is equal to

Answer

To solve the problem step by step, we need to calculate the probability \( k \) that all the dice show different numbers based on the outcome of the coin toss.

### Step 1: Determine the outcomes of the coin toss
A fair coin can either show heads (H) or tails (T). The probability of getting heads or tails is:
\[ P(H) = \frac{1}{2}, \quad P(T) = \frac{1}{2} \]

### Step 2: Calculate the probability for heads (2 dice)
If the coin shows heads, we roll 2 fair dice. We need to find the probability that both dice show different numbers.

- The total number of outcomes when rolling 2 dice is \( 6 \times 6 = 36 \).
- The number of favorable outcomes where both dice show different numbers is \( 6 \) (for the first die) and \( 5 \) (for the second die, which must be different from the first). Thus, the number of favorable outcomes is:
\[ 6 \times 5 = 30 \]
- Therefore, the probability \( P(\text{different numbers | H}) \) is:
\[ P(\text{different numbers | H}) = \frac{30}{36} = \frac{5}{6} \]

### Step 3: Calculate the probability for tails (3 dice)
If the coin shows tails, we roll 3 fair dice. We need to find the probability that all three dice show different numbers.

- The total number of outcomes when rolling 3 dice is \( 6 \times 6 \times 6 = 216 \).
- The number of favorable outcomes where all three dice show different numbers is \( 6 \) (for the first die), \( 5 \) (for the second die), and \( 4 \) (for the third die). Thus, the number of favorable outcomes is:
\[ 6 \times 5 \times 4 = 120 \]
- Therefore, the probability \( P(\text{different numbers | T}) \) is:
\[ P(\text{different numbers | T}) = \frac{120}{216} = \frac{5}{9} \]

### Step 4: Calculate the total probability \( k \)
Now we can calculate the total probability \( k \) that all the dice show different numbers, considering both cases of the coin toss:
\[ k = P(H) \cdot P(\text{different numbers | H}) + P(T) \cdot P(\text{different numbers | T}) \]
Substituting the values:
\[ k = \left(\frac{1}{2} \cdot \frac{5}{6}\right) + \left(\frac{1}{2} \cdot \frac{5}{9}\right) \]
Calculating each term:
\[ k = \frac{5}{12} + \frac{5}{18} \]
To add these fractions, we need a common denominator, which is 36:
\[ k = \frac{5 \cdot 3}{36} + \frac{5 \cdot 2}{36} = \frac{15}{36} + \frac{10}{36} = \frac{25}{36} \]

### Step 5: Calculate \( 180k \)
Now we compute \( 180k \):
\[ 180k = 180 \cdot \frac{25}{36} \]
Calculating this gives:
\[ 180k = 5 \cdot 25 = 125 \]

### Final Answer
Thus, the value of \( 180k \) is:
\[ \boxed{125} \]