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Chapter 7: Problem 52

Human blood is divided into 8 possible blood types. The rarest blood type is AB negative. Only \(1 \%\) of the population has this blood type. Suppose a random sample of 50 people is selected. Can we find the probability that more than \(3 \%\) of the sample has AB negative blood? If so, find the probability. If not, explain why this probability cannot be calculated.

Short Answer

Expert verified
The probability that more than \(3 \%\) (more than 2 people) of the sample has AB negative blood type can be calculated by using the binomial probability formula. Once evaluated, we will get the final value which is the required probability.

Step by step solution

01

Identify the Given Parameters

The problem has given us the following parameters - Probability of success (p) which is the probability that a person has AB negative blood is \(1 % = 0.01\), the sample size (n) which is 50 people, and the number of successes k which is more than \(3 \%\) of sample size that equates to 2 people (since \(3 \%\) of 50 is 1.5, we take the higher integer value which is 2).
02

Apply the Binomial Probability Formula

The task is to find the probability that more than 2 people (i.e., 3 or more people) in the sample have AB negative blood type. Since we need to know about the probability of more than 2 patients, this is equal to one subtracted by the probability up to 2 (P(X 鈮� 2)) occurrences using the binomial cumulative distribution formula which is \( P(X=k) = C(n, k) \times (p)^k \times (1-p)^{(n-k)}\) . Here C(n,k) calculates the number of ways you can choose k successes from n trials. We will have to calculate this for k=0,1, into 2 and add them together to get P(X 鈮� 2). Then, subtract this from 1 to get our desired probability.
03

Compute The Probability

Compute the probabilities for k = 0, 1, and 2 one by one and then add them together. So, Probability that none (k=0) of them has AB negative blood is: \(P(X=0) = C(50, 0) \times (0.01)^0 \times (1-0.01)^{50-0}\). Probability that exactly one (k=1) of them has AB negative blood is: \(P(X=1) = C(50, 1) \times (0.01)^1 \times (1-0.01)^{50-1}\). Probability that exactly two (k=2) of them has AB negative blood is: \(P(X=2) = C(50, 2) \times (0.01)^2 \times (1-0.01)^{50-2}\). Now sum up these three probabilities and subtract it from 1: \(1 - (P(X=0)+P(X=1)+P(X=2))\). Evaluate the expression to compute the probability.

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