Sex ratio at birth is biased towards females in a mongoose population. If the probability of having a daughter is 0.7 in this population, and if sex determination of each offspring is an independent event, then the probability that a female with a litter of four offspring has at least one son is \(\underline{\hspace{0cm}}\) (Round off to two decimal places.)
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Correct Answer: 0.74
Solution and Explanation
Step 1: Understand the problem setup.
The question gives the probability of having a daughter as \( P(\text{daughter}) = 0.7 \). This implies that the probability of having a son is:
\[
P(\text{son}) = 1 - 0.7 = 0.3.
\]
Step 2: Calculate the probability of having no sons.
The probability of having no sons (i.e., all four offspring are daughters) is:
\[
P(\text{no sons}) = (0.7)^4 = 0.2401.
\]
Step 3: Calculate the probability of having at least one son.
The probability of having at least one son is the complement of the probability of having no sons:
\[
P(\text{at least one son}) = 1 - P(\text{no sons}) = 1 - 0.2401 = 0.7599.
\]
Step 4: Conclusion.
Thus, the probability that a female with four offspring has at least one son is approximately \( 0.76 \).
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