5 different games are to be distributed among 4 children randomly. The probability that each child gets atleast one game is

Options

A. `1/4`
B. `15/64`
C. `13/64`
D. `17/64`

Answer

To find the probability that each child gets at least one game when distributing 5 different games among 4 children, we can follow these steps:

### Step 1: Calculate the Total Number of Distributions
Each of the 5 games can be given to any of the 4 children. Therefore, the total number of ways to distribute the 5 games is given by:

\[ \text{Total distributions} = 4^5 \]

Calculating this:

\[ 4^5 = 1024 \]

### Step 2: Calculate the Number of Favorable Distributions
To find the number of ways to distribute the games such that each child gets at least one game, we can use the principle of inclusion-exclusion.

1. **Calculate the total distributions without restrictions:** As calculated above, this is \( 4^5 = 1024 \).

2. **Subtract the cases where at least one child gets no game:**
- Choose 1 child to receive no game. The remaining 3 children can receive the games. The number of ways to distribute the games among 3 children is \( 3^5 \).
- There are \( \binom{4}{1} = 4 \) ways to choose which child receives no game.

So, the total number of distributions where at least one child gets no game is:

\[ 4 \times 3^5 = 4 \times 243 = 972 \]

3. **Add back the cases where at least two children get no game (overcounted):**
- Choose 2 children to receive no game. The remaining 2 children can receive the games. The number of ways to distribute the games among 2 children is \( 2^5 \).
- There are \( \binom{4}{2} = 6 \) ways to choose which 2 children receive no game.

So, the total number of distributions where at least two children get no game is:

\[ 6 \times 2^5 = 6 \times 32 = 192 \]

4. **Subtract the cases where at least three children get no game (overcounted again):**
- Choose 3 children to receive no game. The remaining child must receive all games. There is only 1 way to do this.
- There are \( \binom{4}{3} = 4 \) ways to choose which 3 children receive no game.

So, the total number of distributions where at least three children get no game is:

\[ 4 \times 1^5 = 4 \times 1 = 4 \]

5. **Applying Inclusion-Exclusion:**
Using the principle of inclusion-exclusion, we can find the total number of distributions where at least one child gets no game:

\[ \text{At least one child gets no game} = 972 - 192 + 4 = 784 \]

6. **Calculate the number of favorable distributions:**
The number of distributions where each child gets at least one game is:

\[ \text{Favorable distributions} = \text{Total distributions} - \text{At least one child gets no game} \]

\[ \text{Favorable distributions} = 1024 - 784 = 240 \]

### Step 3: Calculate the Probability
The probability that each child gets at least one game is given by the ratio of the number of favorable distributions to the total distributions:

\[ P(\text{each child gets at least one game}) = \frac{\text{Favorable distributions}}{\text{Total distributions}} = \frac{240}{1024} \]

Simplifying this fraction:

\[ P = \frac{240 \div 16}{1024 \div 16} = \frac{15}{64} \]

### Final Answer
The probability that each child gets at least one game is:

\[ \frac{15}{64} \]