Question

If four calves are born on a given day. What is the chance that two will be males and two will be females?

• A. 3/16
• B. 1/16
• C. 2/16
• D. 6/16

Hint:

For any problem involving birth gender with equal probabilities, the denominator for \( n \) births is always \( 2^n \).
For \( n = 4 \), the denominator is \( 2^4 = 16 \).
The numerator is simply the combination \( \binom{n}{k} \), which for \( \binom{4}{2} \) is \( 6 \), leading instantly to \( 6/16 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The determination of sex for each newborn calf is an independent random event with two possible outcomes.
The probability of a calf being male (\( M \)) is \( p = 1/2 \), and the probability of being female (\( F \)) is \( q = 1/2 \).
Since the births are independent and have constant probabilities, we can model this scenario using the binomial distribution.
Key Formula or Approach:
The binomial probability of obtaining exactly \( k \) successes (males) in \( n \) independent trials (births) is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where:
\( n = 4 \) (total births),
\( k = 2 \) (number of males),
\( p = 1/2 \) (probability of a male),
\( q = 1/2 \) (probability of a female).

Step 2: Detailed Explanation:
Let us substitute the values into the binomial formula to find the exact probability.
The parameters are \( n = 4 \) and \( k = 2 \).
\[ P(\text{2 Males and 2 Females}) = \binom{4}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{4-2} \]
First, we evaluate the binomial coefficient \( \binom{4}{2} \), which gives the number of unique combinations of births:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
Next, we calculate the probability of any single specific birth sequence:
\[ \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^2 = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \]
Now, we multiply the number of combinations by the probability of a single sequence:
\[ P = 6 \times \frac{1}{16} = \frac{6}{16} \]
This gives us a final probability of \( 6/16 \) (which simplifies to \( 3/8 \)).
Since the options are listed as sixteenths, the matching value is \( 6/16 \).

Step 3: Final Answer:
The probability that exactly two calves will be males and two will be females is \( 6/16 \).