Math  /  Data & Statistics

QuestionLouisa Pectol 11/25/24 9:29 PM Question 10, "5.8.33 HW Score: 90%,990 \%, 9 of 10 points rule and trees Part 1 of 2 Points: 0 of 1 Save
In a given population approximately 7%7 \% of all males are color-blind, while only about 0.67%0.67 \% of women are color-blind. Six years ago 48.6%48.6 \% of the population were male and 51.4%51.4 \% were female. (a) What is the probability that a randomly selected person is color-blind? (b) What is the probability that a randomly selected person who is color-blind is female?

Studdy Solution

STEP 1

What is this asking? What are the chances of picking a color-blind person at random, and if we *do* pick a color-blind person, what are the chances that person is a woman? Watch out! Don't mix up the percentages for men and women!
Also, remember we're looking at the whole population, not just men or just women.

STEP 2

1. Calculate the probability of color-blindness for the whole population.
2. Calculate the probability of a color-blind person being female.

STEP 3

Let's **define** some variables to make things easier!
Let MM be the event that a randomly selected person is male, and CC be the event that a randomly selected person is color-blind.
We're given that P(M)=0.486P(M) = 0.486 and P(CM)=0.07P(C|M) = 0.07.
This means that the probability of being male is **0.486**, and the probability of being color-blind *given* that you're male is **0.07**.

STEP 4

Similarly, let FF be the event that a randomly selected person is female.
We know P(F)=0.514P(F) = 0.514 and P(CF)=0.0067P(C|F) = 0.0067.
So, the probability of being female is **0.514**, and the probability of being color-blind *given* that you're female is **0.0067**.

STEP 5

Now, to find the probability of someone being color-blind overall, we need to consider both men and women.
We can use the **law of total probability**: P(C)=P(CM)P(M)+P(CF)P(F)P(C) = P(C|M) \cdot P(M) + P(C|F) \cdot P(F) This formula basically says that the total probability of being color-blind is the sum of the probabilities of being color-blind and male, and being color-blind and female.
Makes sense, right?

STEP 6

Let's plug in the values: P(C)=(0.070.486)+(0.00670.514)P(C) = (0.07 \cdot 0.486) + (0.0067 \cdot 0.514) P(C)=0.03402+0.0034578P(C) = 0.03402 + 0.0034578P(C)=0.0374778P(C) = 0.0374778So, the probability of a randomly selected person being color-blind is approximately **0.0375**.

STEP 7

Now we want to find the probability that a randomly selected color-blind person is female.
This is a *conditional probability*, written as P(FC)P(F|C).
We can use **Bayes' theorem** for this, which is super useful for flipping conditional probabilities: P(FC)=P(CF)P(F)P(C)P(F|C) = \frac{P(C|F) \cdot P(F)}{P(C)}

STEP 8

We already calculated all the values we need! P(CF)=0.0067P(C|F) = 0.0067, P(F)=0.514P(F) = 0.514, and P(C)=0.0374778P(C) = 0.0374778.
Let's plug them in: P(FC)=0.00670.5140.0374778P(F|C) = \frac{0.0067 \cdot 0.514}{0.0374778} P(FC)=0.00344380.0374778P(F|C) = \frac{0.0034438}{0.0374778}P(FC)0.09188P(F|C) \approx 0.09188So, the probability that a randomly selected color-blind person is female is approximately **0.0919**.

STEP 9

(a) The probability that a randomly selected person is color-blind is approximately **0.0375** or **3.75%**. (b) The probability that a randomly selected color-blind person is female is approximately **0.0919** or **9.19%**.

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