What is the probability that exactly three of the five offspring will be brown?

Steps to Solve

1. Identify the probability function

The problem states that we want to find the probability of exactly three brown offspring out of five. The probability mass function for a binomial distribution is given by:

$$ P(r) = \binom{n}{r} p^{r} (1-p)^{n-r} $$

where:

• $n$ = total number of trials (offspring), which is 5.
• $r$ = number of successful outcomes (brown offspring), which is 3.
• $p$ = probability of success on an individual trial (brown offspring), which is given as 0.5.

2. Calculate the binomial coefficient

The binomial coefficient $\binom{n}{r}$ can be calculated using the formula:

$$ \binom{n}{r} = \frac{n!}{r!(n-r)!} $$

For our case where $n = 5$ and $r = 3$:

$$ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4}{2 \times 1} = 10 $$

3. Calculate $p^{r} (1-p)^{n-r}$

Now, using the values defined earlier:

$$ p^{r} = (0.5)^{3} = 0.125 $$

And for the failure probability (not brown):

$$ (1-p)^{n-r} = (0.5)^{5-3} = (0.5)^{2} = 0.25 $$

Now combine these values:

$$ p^{r} (1-p)^{n-r} = 0.125 \times 0.25 = 0.03125 $$

4. Combine all components to find the final probability

Now substitute everything into the probability function:

$$ P(3) = \binom{5}{3} \cdot (0.5)^{3} \cdot (0.5)^{2} $$

So it becomes:

$$ P(3) = 10 \cdot 0.125 \cdot 0.25 = 10 \cdot 0.03125 = 0.3125 $$