What is the probability of having at least two girls? XXX) (Use the form What is the probability of having at most one boy? XXX) (Use the form p What is the probability of having at most one girl? (Use the form XXX) What is the probability of having at most two boys? (Use the form XXX What is the probability of having at most two girls? (Use the form XXX
The Correct Answer and Explanation is:
To answer these probability questions, we need to make assumptions about the family scenario, typically assuming that each child has an equal chance of being a boy or a girl. Let’s assume we are dealing with a family of 3 children and the probability of each child being a boy or a girl is 0.5.
1. Probability of Having at Least Two Girls
The probability of having at least two girls means the family has either 2 or 3 girls. This is calculated by finding the probability of the complementary event (fewer than 2 girls), and then subtracting it from 1. The possible outcomes for the number of girls are 0, 1, 2, or 3. To calculate:
- 0 girls (all boys): The probability is P(0 girls)=(12)3=18P(0 \text{ girls}) = \left( \frac{1}{2} \right)^3 = \frac{1}{8}P(0 girls)=(21)3=81.
- 1 girl: The probability is P(1 girl)=3×(12)3=38P(1 \text{ girl}) = 3 \times \left( \frac{1}{2} \right)^3 = \frac{3}{8}P(1 girl)=3×(21)3=83.
Thus, the probability of having at least two girls is:
P(at least 2 girls)=1−P(0 girls)−P(1 girl)=1−18−38=48=12P(\text{at least 2 girls}) = 1 – P(0 \text{ girls}) – P(1 \text{ girl}) = 1 – \frac{1}{8} – \frac{3}{8} = \frac{4}{8} = \frac{1}{2}P(at least 2 girls)=1−P(0 girls)−P(1 girl)=1−81−83=84=21
2. Probability of Having at Most One Boy
At most one boy means the family has either 0 or 1 boy. This is the complementary event of having more than one boy. We calculate the probability of having 0 or 1 boy:
- 0 boys (all girls): The probability is P(0 boys)=(12)3=18P(0 \text{ boys}) = \left( \frac{1}{2} \right)^3 = \frac{1}{8}P(0 boys)=(21)3=81.
- 1 boy: The probability is P(1 boy)=3×(12)3=38P(1 \text{ boy}) = 3 \times \left( \frac{1}{2} \right)^3 = \frac{3}{8}P(1 boy)=3×(21)3=83.
Thus, the probability of having at most one boy is:
P(at most 1 boy)=P(0 boys)+P(1 boy)=18+38=48=12P(\text{at most 1 boy}) = P(0 \text{ boys}) + P(1 \text{ boy}) = \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}P(at most 1 boy)=P(0 boys)+P(1 boy)=81+83=84=21
3. Probability of Having at Most One Girl
At most one girl means the family has either 0 or 1 girl. This is the complementary event of having more than one girl:
- 0 girls (all boys): The probability is P(0 girls)=(12)3=18P(0 \text{ girls}) = \left( \frac{1}{2} \right)^3 = \frac{1}{8}P(0 girls)=(21)3=81.
- 1 girl: The probability is P(1 girl)=3×(12)3=38P(1 \text{ girl}) = 3 \times \left( \frac{1}{2} \right)^3 = \frac{3}{8}P(1 girl)=3×(21)3=83.
Thus, the probability of having at most one girl is:
P(at most 1 girl)=P(0 girls)+P(1 girl)=18+38=48=12P(\text{at most 1 girl}) = P(0 \text{ girls}) + P(1 \text{ girl}) = \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}P(at most 1 girl)=P(0 girls)+P(1 girl)=81+83=84=21
4. Probability of Having at Most Two Boys
At most two boys means the family has 0, 1, or 2 boys. We can directly sum the probabilities of these events:
- 0 boys (all girls): The probability is P(0 boys)=18P(0 \text{ boys}) = \frac{1}{8}P(0 boys)=81.
- 1 boy: The probability is P(1 boy)=38P(1 \text{ boy}) = \frac{3}{8}P(1 boy)=83.
- 2 boys: The probability is P(2 boys)=3×(12)3=38P(2 \text{ boys}) = 3 \times \left( \frac{1}{2} \right)^3 = \frac{3}{8}P(2 boys)=3×(21)3=83.
Thus, the probability of having at most two boys is:
P(at most 2 boys)=P(0 boys)+P(1 boy)+P(2 boys)=18+38+38=78P(\text{at most 2 boys}) = P(0 \text{ boys}) + P(1 \text{ boy}) + P(2 \text{ boys}) = \frac{1}{8} + \frac{3}{8} + \frac{3}{8} = \frac{7}{8}P(at most 2 boys)=P(0 boys)+P(1 boy)+P(2 boys)=81+83+83=87
5. Probability of Having at Most Two Girls
At most two girls means the family has 0, 1, or 2 girls. We can directly sum the probabilities of these events:
- 0 girls (all boys): The probability is P(0 girls)=18P(0 \text{ girls}) = \frac{1}{8}P(0 girls)=81.
- 1 girl: The probability is P(1 girl)=38P(1 \text{ girl}) = \frac{3}{8}P(1 girl)=83.
- 2 girls: The probability is P(2 girls)=3×(12)3=38P(2 \text{ girls}) = 3 \times \left( \frac{1}{2} \right)^3 = \frac{3}{8}P(2 girls)=3×(21)3=83.
Thus, the probability of having at most two girls is:
P(at most 2 girls)=P(0 girls)+P(1 girl)+P(2 girls)=18+38+38=78P(\text{at most 2 girls}) = P(0 \text{ girls}) + P(1 \text{ girl}) + P(2 \text{ girls}) = \frac{1}{8} + \frac{3}{8} + \frac{3}{8} = \frac{7}{8}P(at most 2 girls)=P(0 girls)+P(1 girl)+P(2 girls)=81+83+83=87
Summary
- At least two girls: 12\frac{1}{2}21
- At most one boy: 12\frac{1}{2}21
- At most one girl: 12\frac{1}{2}21
- At most two boys: 78\frac{7}{8}87
- At most two girls: 78\frac{7}{8}87
