Find the probability of at least 2 girls in 10 births. Assume that male and female births are equally likely and that the births are independent events.
Well, let's start out by figuring out the probability of having all boys or all girls in 10 births. Each birth has a 50% chance of being a girl and a 50% chance of being a boy. So the probability of having all girls is (0.5)^10, and the probability of having all boys is also (0.5)^10.
The probability of having at least 2 girls in 10 births is the complement of having all boys or all girls. In other words, it's 1 minus the probability of having all boys or all girls.
So the probability of having at least 2 girls in 10 births is 1 - [(0.5)^10 + (0.5)^10] = 1 - 2*(0.5)^10.
Now, let's calculate that:
1 - 2*(0.5)^10 = 1 - 2*(1/1024) = 1 - 2/1024 = 1 - 1/512.
So the probability of having at least 2 girls in 10 births is approximately 1 - 1/512.
But just remember, probabilities are just like that one friend who texts you first thing in the morning, they can be a bit unpredictable!
To find the probability of at least 2 girls in 10 births, we can use the concept of binomial probability and complement rule.
The probability of having a girl in a single birth is 0.5, assuming male and female births are equally likely. Similarly, the probability of having a boy in a single birth is also 0.5.
The number of successes (girls) in n independent trials (births) follows a binomial distribution with parameters n and p, where n is the number of trials and p is the probability of success (having a girl).
To find the probability of at least 2 girls, we need to sum up the probabilities of having 2, 3, 4, ..., 10 girls.
P(at least 2 girls) = 1 - P(no girls) - P(1 girl)
P(no girls) = (0.5)^10, as the probability of having no girls in 10 births is (0.5)^10
P(1 girl) = C(10,1) * (0.5)^1 * (0.5)^9, as the probability of having only 1 girl in 10 births is equal to the number of combinations of choosing 1 from 10 multiplied by the probability of having 1 girl and 9 boys.
P(at least 2 girls) = 1 - (0.5)^10 - C(10,1) * (0.5)^1 * (0.5)^9
Now, let's calculate this probability step by step:
P(no girls) = (0.5)^10 = 0.0009765625
P(1 girl) = C(10, 1) * (0.5)^1 * (0.5)^9 = 10 * 0.5 * (0.5)^9 = 0.009765625
P(at least 2 girls) = 1 - (0.0009765625) - 0.009765625 = 0.9892578125
Therefore, the probability of having at least 2 girls in 10 births is approximately 0.9893, or 98.93%.
To find the probability of at least 2 girls in 10 births, we can use the binomial probability formula and calculate the probability of each possible outcome and then sum them up.
First, let's calculate the probability of having exactly 2 girls in 10 births:
For each birth, the probability of having a girl is 1/2, and the probability of having a boy is also 1/2.
Using the binomial probability formula, the probability of having exactly 2 girls in 10 births is:
P(X = 2) = (10 C 2) * (1/2)^2 * (1/2)^(10-2)
Where:
(10 C 2) represents the number of ways to choose 2 girls from 10 births, which is calculated as 10! / (2! * (10-2)!)
(1/2)^2 represents the probability of having 2 girls
(1/2)^(10-2) represents the probability of having 8 boys
You can calculate this using a calculator or manually, using the formula above.
Next, let's calculate the probability of having exactly 3 girls, 4 girls, and so on, up to 10 girls, and sum them up.
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + ... + P(X = 10)
This will give us the probability of having at least 2 girls in 10 births.
Note: The above calculation assumes that male and female births are equally likely, and that the births are independent events.
The question satisfies requirements for a binomial distribution, where
n=10
p=1/2
Probability of r girls out of n
=P(G=r)
=(n, r)p^r (1-p)^(n-r)
where
(n,r)=combination n choose r
= n!/(r!(n-r)!)
For at least 2 girls out of 10,
P(G>=2)=1-(P(G=0)+P(G=1))
1-(1)(1)(1/1024)-(10)(1/2)(1/512)
=1-11/1024
=1013/1024