Assuming that both parents are heterozygous for a recessive gene, the probability that a single offspring develops the recessive trait is 1/4. Assuming separate births are independent, the probability that exactly three out of four offspring develop the trait is given by

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This is a binomial probability problem where the probability of success (in this case, an offspring developing the recessive trait) is $ \frac{1}{4} $, and the probability of failure (not developing the trait) is $ \frac{3}{4} $. The formula for exactly $k$ successes in $n$ trials is $ \binom{n}{k} p^k (1-p)^{n-k} $, where $p$ is the probability of success. For exactly 3 out of 4 offspring developing the trait, $n=4$, $k=3$, and $p=\frac{1}{4}$, so the probability is $ \binom{4}{3} \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^{4-3} = 4 \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^1 $.

Assuming That Both Parents Are Heterozygous for a Recessive Gene 4(14)3(34)14 \left( \frac { 1 } { 4 } \right) ^ { 3 } \left( \frac { 3 } { 4 } \right) ^ { 1 }4(41​)3(43​)1

Assuming That Both Parents Are Heterozygous for a Recessive Gene $4 \left( \frac { 1 } { 4 } \right) ^ { 3 } \left( \frac { 3 } { 4 } \right) ^ { 1 }$