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About 3 out of 100 mothers give birth to twins. If three mothers are randomly selected from this population, what is the probability that at least one of the three mothers did NOT give birth to twins? Responses A 0.9126730.912673 B 0.9152830.915283 C 0.9973540.997354 D 0.999973

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Final answer:

To calculate the probability that at least one of three selected mothers did not give birth to twins, we subtract the complement event (all having twins) from 1. The probability is approximately 0.999973.

Step-by-step explanation:

The subject question involves calculating the probability of an event in a population of mothers giving birth to twins. We know that 3 out of 100 or 0.03 mothers have twins. The probability that a mother does NOT have twins is then 1 - 0.03 = 0.97. To find the probability that at least one of three randomly selected mothers did not give birth to twins, we consider the complementary event: all three mothers having twins. The probability of this event is (0.03)^3.

To compute the desired probability, we subtract the probability of the complement from 1: 1 - (0.03)^3 = 0.999973. Hence, the probability of at least one mother not having twins is approximately 0.999973.

The complementary probability is the probability of the event not happening. In this case, the complementary event is that all three selected mothers gave birth to twins. The probability of a mother giving birth to twins is 3/100, so the probability of all three selected mothers giving birth to twins is (3/100)^3.

Therefore, the probability that at least one of the three mothers did not give birth to twins is 1 - (3/100)^3.

Simplifying this expression, we get a probability of approximately 0.999973, which is option D.

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