Suppose the Internal Revenue Service is studying the category of charitable contributions. A sample of 26 returns is selected from young couples between the ages of 20 and 35 who had an adjusted gross income of more than $100,000. Of these 26 returns 6 had charitable contributions of more than $1,000. Suppose 5 of these returns are selected for a comprehensive audit.
What is the probability exactly one of the five audited had a charitable deduction of more than $1,000? (Round your answer to 4 decimal places.)
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To solve this problem, we can use the concept of probability and the binomial distribution. The binomial distribution is used to calculate the probability of a specific number of successes in a fixed number of independent Bernoulli trials.
In this case, the probability of each return having a charitable contribution of more than $1,000 can be considered a Bernoulli trial. There are 26 total returns, of which we know 6 had charitable contributions of more than $1,000. The probability of success (p) for each trial is 6/26 = 0.2308.
We are asked to find the probability of exactly one of the five audited returns having a charitable deduction of more than $1,000. This means we want to calculate P(X = 1), where X follows a binomial distribution with parameters n = 5 (number of trials) and p = 0.2308 (probability of success).
The formula for the probability mass function of the binomial distribution is:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where (n choose k) is the binomial coefficient, defined as (n! / (k! * (n-k)!)).
Plugging in the values, we have:
P(X = 1) = (5 choose 1) * (0.2308^1) * (1-0.2308)^(5-1)
Calculating the binomial coefficient:
(5 choose 1) = 5! / (1! * (5-1)!) = 5
Plugging in the values:
P(X = 1) = 5 * (0.2308^1) * (1-0.2308)^(5-1)
Calculating the result:
P(X = 1) = 5 * 0.2308 * (1-0.2308)^4
P(X = 1) ≈ 0.0142 (rounded to 4 decimal places)
Therefore, the probability exactly one of the five audited returns had a charitable deduction of more than $1,000 is approximately 0.0142.