Question

probabilities of blood types of blood types a, b, e rh factor that is either use the information in the two - way table to comp the statement. the probability that a person has a positive rh factor given that he/she has type o blood is percent. there is a greater probability for a person to have a positive rh factor given type o blood.

Explanation:

Step1: Identify relevant probabilities

We need the probability of positive Rh factor given type O blood. Let \( P(+) \) be positive Rh factor, \( P(O) \) be type O blood, and \( P(+|O) \) be the conditional probability. From the table, the probability of type O and positive Rh factor is \( 0.37 \), and the probability of type O blood (total for O) is \( 0.45 \) (wait, no, looking at the table: the O column, total? Wait, the table has rows and columns. Let's re-express. The two - way table: for blood type O, the positive Rh factor probability (the cell for O and positive) is \( 0.37 \)? Wait, no, let's parse the table. The columns are blood types (A, B, AB, O) and rows are Rh factor (positive, negative) maybe? Wait, the total for O blood: looking at the "Total" column for O, the total probability of O blood is \( 0.45 \)? Wait, no, the bottom row is total, with 1.0. Wait, the O column: the positive Rh factor cell (assuming rows are positive/negative) – let's see, the O column has a cell with \( 0.37 \) (positive) and \( 0.08 \) (negative)? Wait, no, the table as given: let's list the cells. For O blood: the probability of O and positive Rh factor is \( 0.37 \), and the probability of O blood (marginal probability) is \( 0.37 + 0.08=0.45 \)? Wait, no, the "Total" row for O is \( 0.45 \)? Wait, the bottom row: A total is \( 0.40 \), B total is \( 0.11 \), AB total is \( 0.04 \), O total is \( 0.45 \), and total is \( 1.0 \). The positive Rh factor for O: looking at the O column, the positive cell is \( 0.37 \), negative is \( 0.08 \). So \( P(O\cap +)=0.37 \), \( P(O) = 0.45 \).

Step2: Apply conditional probability formula

The formula for conditional probability is \( P(+|O)=\frac{P(O\cap +)}{P(O)} \)

So \( P(+|O)=\frac{0.37}{0.45}\approx0.8222 \), or \( 82.22\% \). Wait, but the other part: the probability of positive Rh factor in general? Wait, no, the question is: "the probability that a person has a positive Rh factor given that he/she has type O blood". So using conditional probability: \( P(+|O)=\frac{P(O\cap +)}{P(O)} \)

From the table, \( P(O\cap +) = 0.37 \), \( P(O)=0.37 + 0.08 = 0.45 \)? Wait, no, the "Total" for O blood is \( 0.45 \) (from the bottom row: O total is \( 0.45 \)). And the positive Rh factor for O is \( 0.37 \), negative is \( 0.08 \) (since \( 0.37+0.08 = 0.45 \)).

So \( P(+|O)=\frac{0.37}{0.45}\approx0.8222 \), or \( 82.22\% \).

Wait, but also, the other part: "There is a greater probability for a person to have a positive Rh factor given type O blood" – wait, maybe we also need to calculate the probability of positive Rh factor in general? Wait, no, the question is about conditional probability. Wait, the first part: "the probability that a person has a positive Rh factor given that he/she has type O blood".

Step1 (corrected):

Let's re - examine the table. The two - way table:

• Probability of O and positive (\( P(O\cap +) \)): \( 0.37 \)
• Probability of O blood (\( P(O) \)): \( 0.37 + 0.08=0.45 \) (since the O column has positive \( 0.37 \) and negative \( 0.08 \), and total for O is \( 0.45 \))

Step2 (corrected):

Using the conditional probability formula \( P(+|O)=\frac{P(O\cap +)}{P(O)}=\frac{0.37}{0.45}\approx0.8222 \), which is \( 82.22\% \)

Now, for the other part: "There is a greater probability for a person to have a positive Rh factor given type O blood" – we can compare with other blood types, but the main calculation is for \( P(+|O) \)

Answer:

The probability that a person has a positive Rh factor given that he/she has type O blood is \( \frac{37}{45}\approx0.8222 \) or \( 82.22\% \) (if we consider the calculation \( \frac{0.37}{0.45}\approx0.822 \)). If we made a mistake in table parsing, let's re - check. Wait, maybe the O column's positive is \( 0.37 \) and O total is \( 0.37 + 0.08 = 0.45 \), so the conditional probability is \( \frac{0.37}{0.45}\approx82.2\% \)