an insurance company reported that 55% of all automobile damage claims were made by people under age 25. If six automobile damage claims were selected at random, determine the probility that exactly four of them were made by someone under ager 25.
Well, deciding the probability of such a specific scenario requires a bit of math, so let's do some clown calculations!
First, let's break it down step by step. We know that 55% of all claims are made by people under 25. So, the probability of one claim being made by someone under 25 is 0.55.
Now we need to consider the probability of exactly four out of six claims being made by someone under 25. We can calculate this using the binomial distribution formula:
P(X=k) = (nCk) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of exactly k successes (in this case, 4 claims made by someone under 25).
n is the total number of trials (in this case, 6 claims).
k is the number of desired successes (in this case, 4 claims made by someone under 25).
p is the probability of a single success (in this case, 0.55).
Let's plug in the numbers and solve:
P(X=4) = (6C4) * 0.55^4 * (1-0.55)^(6-4)
Calculating the combination (6C4) using the formula nCr = n! / (r!(n-r)!), we get:
P(X=4) = (6! / (4!(6-4)!)) * 0.55^4 * 0.45^2
After doing the math, we find that the probability of exactly four out of six claims being made by someone under 25 is approximately 0.272, or 27.2%.
Remember, these calculations rely on the assumption that each claim is independent of one another and that the 55% statistic holds true. So, don't take these numbers too seriously!
To solve this problem, we can use the binomial probability formula:
P(X = k) = (n C k) * p^k * q^(n-k)
Where:
P(X = k) is the probability of getting exactly k successes
n is the number of trials
k is the number of successes
p is the probability of success
q is the probability of failure (1 - p)
n C k is the combination formula: n! / (k! * (n-k)!)
In this case:
n = 6 (number of automobile damage claims selected)
k = 4 (number of claims made by someone under age 25)
p = 0.55 (probability of a claim made by someone under age 25)
q = 0.45 (probability of a claim made by someone 25 or older)
Plugging in these values, we can calculate the probability:
P(X = 4) = (6 C 4) * 0.55^4 * 0.45^(6-4)
Calculating the combination:
(6 C 4) = 6! / (4! * (6-4)!) = 6! / (4! * 2!) = (6 * 5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * (2 * 1)) = 15
Plugging in the values into the formula:
P(X = 4) = 15 * (0.55^4) * (0.45^2)
Calculating:
P(X = 4) = 15 * 0.09150625 * 0.2025
P(X = 4) ≈ 0.1386
Therefore, the probability that exactly four out of six automobile damage claims were made by someone under age 25 is approximately 0.1386 or 13.86%.
To determine the probability of exactly four out of six selected automobile damage claims being made by someone under the age of 25, we can use the binomial probability formula.
The binomial probability formula is given by:
P(X = k) = (nCk) * p^k * (1 - p)^(n-k)
Where:
P(X = k) represents the probability of getting k successes (in this case, four claims made by someone under the age of 25),
n represents the total number of trials or selections (in this case, six claims),
k represents the specific number of successes we want (in this case, four claims made by someone under the age of 25),
p represents the probability of success on a single trial (in this case, the probability of an automobile damage claim made by someone under the age of 25, which is given as 55% or 0.55), and
nCk represents the number of combinations of n items taken k at a time.
To calculate the probability, we can substitute the given values into the formula:
P(X = 4) = (6C4) * (0.55)^4 * (1 - 0.55)^(6 - 4)
Now, let's calculate each component of the formula:
(6C4) = (6!)/(4!(6-4)!) = 6! / (4! * 2!) = (6 * 5 * 4!) / (4! * 2 * 1) = 15
(0.55)^4 = 0.096125
(1 - 0.55)^(6 - 4) = 0.2975
Therefore, the probability of exactly four out of the six selected automobile damage claims being made by someone under the age of 25 is:
P(X = 4) = 15 * 0.096125 * 0.2975 = 0.4269875
So, the probability is approximately 0.427, or 42.7%.
prob(under 25) = .55
prob(over 25) = .45
prob (exactly 4 of 6 are under 25)
= C(6,4) (.55)^4 (.45)^2
= .27795