Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gender of any other children. Find the probability of exactly eight in ten births.
Use the binomial probability function. If you use a binomial probability table, this problem will be easily solved. In most tables, you will need x, n, and p. x = 8, n = 10, and p = .5 (look up the probability in the table using those values).
I hope this will help.
10C8 (0.5)^8 x (0.5)^2 = 0.044
n=10 , x=8 , p = 0.5
out of 10, you choose 8
p of getting boys = 0.5
p of getting girls = 0.5
otherway, get exactly 2 girls
10C2 (0.5)^10 = 0.044
Well, I must say, trying to predict the gender of babies is like trying to predict what flavor of ice cream you'll get with your eyes closed - it's a bit of a guessing game, but let's give it a humorous shot!
If we assume that each birth has an equal chance of being a boy or a girl, we can treat it like a series of independent coin flips (though I wouldn't recommend flipping babies). Since the probability of a boy or girl remains the same for each birth, the probability of any particular combination of births will depend on the number of boys and girls involved.
Now, to find the probability of exactly eight in ten births, we need to consider all the ways we can arrange eight boys and two girls. One possible arrangement is BBBBGGGGGG. But wait, there are more!
Using a bit of counting magic, we find that there are 10 choose 8 (10C8) possible arrangements of eight boys and two girls, which is equal to 45. Why? Because it's like picking 8 out of the 10 births to be boys.
Since each arrangement has an equal chance of occurring, we divide the number of successful arrangements by the total number of possible arrangements. So, the probability of exactly eight in ten births is 45 divided by the total number of possibilities, which is 2 to the power of 10 (since each birth has two possibilities - boy or girl).
Therefore, the probability of exactly eight in ten births is 45/1024, which simplifies to roughly 0.04395, or about 4.4%.
So, there you have it! The probability of exactly eight in ten births being boys or girls (as long as those babies don't have any wild plans to mess with probability) is about 4.4%. Remember, though, when it comes to actual births - nature often loves to throw curveballs, like twins, triplets, or shooting for a perfect gender balance.
To find the probability of exactly eight out of ten births being a certain gender, we can use the concept of binomial probability.
First, let's define the probability of a single birth being the desired gender. Since male and female births are equally likely, the probability of a single birth being male or female is both 1/2 or 0.5.
To find the probability of exactly eight out of ten births being the desired gender, we need to calculate the probability of having eight births of the desired gender and two births of the opposite gender.
Now, we can use the formula for binomial probability:
P(X=k) = (nCk) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of exactly k successes (eight in this case),
n is the total number of trials (ten in this case),
k is the number of desired outcomes (eight in this case),
p is the probability of a single desired outcome (0.5 in this case), and
(nCk) is the number of combinations of n items taken k at a time.
Let's calculate the probability:
P(X=8) = (10C8) * (0.5)^8 * (1-0.5)^(10-8)
Using the combination formula (nCk) = n! / (k! * (n-k)!), we calculate:
P(X=8) = (10! / (8!(10-8)!)) * (0.5)^8 * (1-0.5)^2
Simplifying the combination:
P(X=8) = (10! / (8!2!)) * (0.5)^8 * (1-0.5)^2
Now, we can calculate the probability:
P(X=8) = (10 * 9 / 2 * 1) * (0.5)^8 * (0.5)^2
Simplifying:
P(X=8) = (45) * (0.5)^8 * (0.5)^2
P(X=8) = 45 * 0.00390625 * 0.25
P(X=8) = 0.0439453125
Therefore, the probability of exactly eight out of ten births being the desired gender is approximately 0.044 or 4.4%.