Question

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.)

Hint:

To calculate the probability of having at least one son, subtract the probability of having no sons from 1. This technique is useful when dealing with complementary events.

Answer 1

Correct Answer: 0.74

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 \( 0.76 \).