90%of a countrys population are a right handed . what is the probability that exactly 29 people in a group of 30 are right handad?
Binomial distribution
p(success) = .9
1-p = .1
P(29 out of 30)
= C(30,29) * .9^29 * .1^1
C(30,29) = 30! / [ 1! (29!) ]
= 30
so
30 * .9^29 * .1 = .1413
To find the probability of exactly 29 people out of a group of 30 being right-handed when 90% of the country's population is right-handed, we can use the binomial probability formula.
The formula for calculating the probability of exactly x successes in n trials is:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where:
- n is the total number of trials (number of people in the group)
- x is the number of successes (number of right-handed people)
- p is the probability of success (proportion of right-handed people in the population)
- nCx represents the number of ways to choose x items from a set of n items (combination).
In this case, n = 30, x = 29, and p = 0.9 since 90% of the country's population is right-handed.
Plugging in these values into the formula, we get:
P(29) = (30C29) * (0.9)^29 * (1-0.9)^(30-29)
The combination (nCx) can be calculated as:
nCx = n! / (x!(n-x)!)
Let's calculate each part step-by-step:
1. Calculate the combination (30C29):
30C29 = 30! / (29!(30-29)!) = 30
2. Calculate the probability of having 29 right-handed people:
P(29) = 30 * (0.9)^29 * (1-0.9)^(30-29)
P(29) = 30 * (0.9)^29 * (0.1)^1
3. Evaluate the exponential part:
(0.9)^29 = 0.082
(0.1)^1 = 0.1
4. Multiply all parts together:
P(29) = 30 * 0.082 * 0.1
P(29) = 0.246
Therefore, the probability that exactly 29 people in a group of 30 are right-handed is approximately 0.246, or 24.6%.
To find the probability that exactly 29 people in a group of 30 are right-handed, we need to use the concept of binomial probability.
In this case, the probability of each individual being right-handed is 90% or 0.9 (since 90% of the population are right-handed). This means the probability of someone being left-handed is 10% or 0.1.
The binomial probability formula is:
P(X = k) = nCk * p^k * q^(n-k)
Where:
- P(X = k) is the probability of exactly k successes (in this case, exactly 29 right-handed people)
- n is the total number of trials (in this case, the group size of 30)
- k is the number of successes (in this case, 29)
- p is the probability of success (in this case, 0.9)
- q is the probability of failure (in this case, 0.1)
Now, plugging in the values:
P(X = 29) = 30C29 * 0.9^29 * 0.1^1
The notation "30C29" represents the number of combinations of 30 items taken 29 at a time, which can be calculated as:
30C29 = 30! / (29! * (30-29)!)
Simplifying further:
P(X = 29) = 30 * 0.9^29 * 0.1
Calculating this value will give you the probability that exactly 29 people in a group of 30 are right-handed.