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In a colony all families have at least one child. The probability that a randomly chosen family from this colony has exactly k children is (0.5)k ; k = 1, 2,K . A child is either a male or a female with equal probability. The probability that such a family consists of at least one male child and at least one female child is _________
Correct answer is '0.3 to 0.4'. Can you explain this answer?
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In a colony all families have at least one child. The probability that...
The probability that a randomly chosen family from this colony has exactly k children is given by (0.5)^k, where k = 1, 2, K.
To find the probability that such a family consists of at least one male child and at least one female child, we need to consider two cases:
Case 1: Only one child in the family
Since a family must have at least one child, the probability of having exactly one child is 1. In this case, the probability of having both a male and a female child is 0.5 * 0.5 = 0.25.
Case 2: More than one child in the family
In this case, we need to find the probability of having at least one male child and at least one female child among the k children.
To do this, we can find the probability of the complementary event, which is having either all male or all female children.
The probability of having all male children is (0.5)^k, and the probability of having all female children is also (0.5)^k.
So, the probability of having either all male or all female children is 2 * (0.5)^k.
Therefore, the probability of having at least one male and at least one female child among the k children is 1 - 2 * (0.5)^k.
To find the overall probability, we need to consider both cases.
The probability of having exactly one child is 1, and the probability of having both a male and a female child is 0.25.
So, the overall probability is 1 * 0.25 + (1 - 0.25) * (1 - 2 * (0.5)^k).
Simplifying this expression, we get 0.25 + (1 - 0.25) * (1 - 2 * (0.5)^k) = 0.25 + 0.75 * (1 - 2 * (0.5)^k).
Since the given answer is '0.3 to 0.4', we can infer that the value of 0.25 + 0.75 * (1 - 2 * (0.5)^k) lies within this range for k = 1, 2, K.
To find the probability that such a family consists of at least one male child and at least one female child, we need to consider two cases:
Case 1: Only one child in the family
Since a family must have at least one child, the probability of having exactly one child is 1. In this case, the probability of having both a male and a female child is 0.5 * 0.5 = 0.25.
Case 2: More than one child in the family
In this case, we need to find the probability of having at least one male child and at least one female child among the k children.
To do this, we can find the probability of the complementary event, which is having either all male or all female children.
The probability of having all male children is (0.5)^k, and the probability of having all female children is also (0.5)^k.
So, the probability of having either all male or all female children is 2 * (0.5)^k.
Therefore, the probability of having at least one male and at least one female child among the k children is 1 - 2 * (0.5)^k.
To find the overall probability, we need to consider both cases.
The probability of having exactly one child is 1, and the probability of having both a male and a female child is 0.25.
So, the overall probability is 1 * 0.25 + (1 - 0.25) * (1 - 2 * (0.5)^k).
Simplifying this expression, we get 0.25 + (1 - 0.25) * (1 - 2 * (0.5)^k) = 0.25 + 0.75 * (1 - 2 * (0.5)^k).
Since the given answer is '0.3 to 0.4', we can infer that the value of 0.25 + 0.75 * (1 - 2 * (0.5)^k) lies within this range for k = 1, 2, K.
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In a colony all families have at least one child. The probability that...
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In a colony all families have at least one child. The probability that a randomly chosen family from this colony has exactly k children is (0.5)k ; k = 1, 2,K . A child is either a male or a female with equal probability. The probability that such a family consists of at least one male child and at least one female child is _________ Correct answer is '0.3 to 0.4'. Can you explain this answer?
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