if 1o% of a popuation are tested positive with hiv and 6 people were tested calculate ;[1] probability that exactly 5 will be hiv positive using the shortest possible method?

Aliyuus2

7 months ago

The probability that exactly 5 out of the 6 people tested will be HIV positive is 0.000054 or \(5.4\times 10^{-5}\). Step 1: Identify the Binomial Parameters This problem uses the binomial probability distribution because there are a fixed number of independent trials (people tested), each with two possible outcomes (positive or negative), and a constant probability of success. Number of trials, \(n=6\)Probability of success (HIV positive), \(p=0.1\)Probability of failure (HIV negative), \(q=1-p=0.9\)Desired number of successes, \(k=5\) Step 2: Apply the Binomial Probability Formula The formula to calculate the probability of exactly \(k\) successes in \(n\) trials is:\(P(X=k)={n \choose k}p^{k}(1-p)^{n-k}\)Substituting the values for this problem:\(P(X=5)={6 \choose 5}(0.1)^{5}(0.9)^{6-5}\)\(P(X=5)=6\times (0.00001)\times (0.9)\)\(P(X=5)=0.000054\) Answer: The probability is 0.000054. To explore more scenarios (e.g., probability of at most 2 positives, or none positive), shall we calculate another probability for a different number of positive results?