Out of 800 families with 4 children each, the expected number of families having atleast one boy is

Options

A. 550
B. 50
C. 750
D. 300

Explanation

To solve the problem of finding the expected number of families having at least one boy out of 800 families with 4 children each, we can follow these steps:

### Step 1: Understand the scenario
We have 800 families, each with 4 children. We need to find the expected number of families that have at least one boy.

### Step 2: Define the probability of having no boys
The probability of having a boy or a girl is equal, which is \( p = \frac{1}{2} \) for a boy and \( q = \frac{1}{2} \) for a girl.

The probability of having no boys (i.e., all girls) in one family with 4 children can be calculated using the binomial distribution formula.

The probability of having 0 boys in 4 children is given by:
\[ P(X = 0) = \binom{n}{k} p^k q^{n-k} \]
where \( n = 4 \) (the number of children), \( k = 0 \) (the number of boys), \( p = \frac{1}{2} \), and \( q = \frac{1}{2} \).

Thus, we have:
\[ P(X = 0) = \binom{4}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^4 = 1 \cdot 1 \cdot \left(\frac{1}{16}\right) = \frac{1}{16} \]

### Step 3: Calculate the probability of having at least one boy
The probability of having at least one boy is the complement of having no boys:
\[ P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{1}{16} = \frac{15}{16} \]

### Step 4: Calculate the expected number of families with at least one boy
Now, we multiply the probability of having at least one boy by the total number of families:
\[ \text{Expected number of families with at least one boy} = 800 \times P(X \geq 1) = 800 \times \frac{15}{16} \]

Calculating this gives:
\[ 800 \times \frac{15}{16} = 800 \times 0.9375 = 750 \]

### Final Answer
The expected number of families having at least one boy is **750**.

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