= Solution
Assuming the egg colours are <independent> with the same probability $p$, the number of white eggs has the <binomial distribution> with four trials. The possibilities are
|| Contents
|| Probability
| $W_0B_4$
| $(1-p)^4$
| $W_1B_3$
| $4p(1-p)^3$
| $W_2B_2$
| $6p^2(1-p)^2$
| $W_3B_1$
| $4p^3(1-p)$
| $W_4B_0$
| $p^4$
The <binomial coefficients> count which eggs are white. Marginal colour probabilities alone would not determine nest probabilities without the <independence> assumption.
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