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Three people are chosen at random. (Assume the choices are independent events). What is the probability that they all have the same blood type?

Question

Three people are chosen at random. (Assume the choices are independent events). What is the probability that they all have the same blood type?

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Solution

To answer this question, we need to know the probability of each blood type. There are 8 blood types: A+, A-, B+, B-, AB+, AB-, O+, and O-. Assuming each blood type is equally likely, each person has a 1/8 chance of having a specific blood type.

Step 1: The first person can have any blood type, so the probability is 1.

Step 2: The second person must have the same blood type as the first. Since there's a 1/8 chance of this happening, the probability is 1/8.

Step 3: The third person must also have the same blood type. Again, there's a 1/8 chance of this happening, so the probability is 1/8.

To find the total probability, we multiply the probabilities together: 1 * 1/8 * 1/8 = 1/64.

So, the probability that all three people chosen at random have the same blood type is 1/64, assuming each blood type is equally likely.

Please note that in reality, blood types are not equally distributed in the population. The actual probability would depend on the prevalence of each blood type in the population from which the people are chosen.

This problem has been solved

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