Question

a) the probability that a woman tests positive given that she has breast cancer 7/8
b) the probability that a woman tests positive given that she does not have breast cancer 70/992
determine where or not the events of having breast cancer and testing positive for breast cancer are independent. show all relevant calculations. write answers as fractions.
$p(a \text{ and } b) = p(a) \cdot p(b)$(used to determine independence of events)
c) what is the probability of having breast cancer? 1/125
d) what is the probability of testing positive for breast cancer? 77/1000
e) what is the probability of having breast cancer and testing positive? 7/8
f) are the events of having breast cancer and testing positive for breast cancer independent? use the work from problems c, d, and e to help you answer.
type in yes or no. type your answer..
g) find the probability that a woman has breast cancer given that her test result is positive. type your answer..
h) find the probability that a woman does not have breast cancer given that her test result is negative. type your answer..

Explanation:

Step1: Recall Independence Formula

For events \( A \) (having breast cancer) and \( B \) (testing positive), they are independent if \( P(A \cap B) = P(A) \cdot P(B) \).
We have \( P(A)=\frac{1}{125} \), \( P(B)=\frac{77}{1000} \), and \( P(A \cap B)=\frac{7}{8} \)? Wait, no, wait—wait, actually, let's recalculate \( P(A \cap B) \) correctly. Wait, \( P(A \cap B)=P(B|A) \cdot P(A) \). From part a, \( P(B|A)=\frac{7}{8} \), \( P(A)=\frac{1}{125} \), so \( P(A \cap B)=\frac{7}{8} \times \frac{1}{125}=\frac{7}{1000} \). Wait, the given \( P(A \cap B) \) in e) was wrong? Wait, no, maybe I misread. Wait, let's check the values again.

Wait, let's start over. Let \( A \) be "having breast cancer", \( B \) be "testing positive".

From part c: \( P(A)=\frac{1}{125} \)

From part d: \( P(B)=\frac{77}{1000} \)

From part a: \( P(B|A)=\frac{7}{8} \), so \( P(A \cap B)=P(B|A) \cdot P(A)=\frac{7}{8} \times \frac{1}{125}=\frac{7}{1000} \)

Now calculate \( P(A) \cdot P(B)=\frac{1}{125} \times \frac{77}{1000}=\frac{77}{125000} \approx 0.000616 \)

And \( P(A \cap B)=\frac{7}{1000}=0.007 \)

Since \( \frac{7}{1000} eq \frac{1}{125} \times \frac{77}{1000} \) (because \( \frac{1}{125} \times \frac{77}{1000}=\frac{77}{125000}=\frac{77\div 11}{125000\div 11}=\frac{7}{11363.63...} \) no, wait, \( \frac{1}{125}=0.008 \), \( \frac{77}{1000}=0.077 \), so \( 0.008 \times 0.077 = 0.000616 \), and \( P(A \cap B)=\frac{7}{1000}=0.007 \), which is not equal. So they are not independent.

Wait, but let's check the values again. Wait, maybe the value in e) was a typo? Wait, the user's e) says \( 7/8 \), but that's wrong. Wait, no—wait, maybe I misread the problem. Wait, the problem says "Write answers as fractions". Let's recalculate \( P(A \cap B) \):

\( P(A)=\frac{1}{125} \), \( P(B|A)=\frac{7}{8} \), so \( P(A \cap B)=P(B|A) \times P(A)=\frac{7}{8} \times \frac{1}{125}=\frac{7}{1000} \). So the value in e) should be \( \frac{7}{1000} \), not \( \frac{7}{8} \). That was a mistake. Then, \( P(A) \times P(B)=\frac{1}{125} \times \frac{77}{1000}=\frac{77}{125000} \), and \( P(A \cap B)=\frac{7}{1000}=\frac{875}{125000} \). Since \( \frac{77}{125000} eq \frac{875}{125000} \), the events are not independent.

Step2: Check f)

Using the correct \( P(A \cap B)=\frac{7}{1000} \), \( P(A) \cdot P(B)=\frac{1}{125} \times \frac{77}{1000}=\frac{77}{125000} \). Since \( \frac{7}{1000}=\frac{875}{125000} \) and \( \frac{77}{125000} eq \frac{875}{125000} \), the events are not independent. So the answer is no.

Step3: Calculate g) (Conditional Probability)

We need \( P(A|B)=\frac{P(A \cap B)}{P(B)} \). We have \( P(A \cap B)=\frac{7}{1000} \), \( P(B)=\frac{77}{1000} \), so \( P(A|B)=\frac{7/1000}{77/1000}=\frac{7}{77}=\frac{1}{11} \).

Step4: Calculate h) (Conditional Probability)

Let \( A' \) be "not having breast cancer", \( B' \) be "testing negative". We know \( P(A')=1 - P(A)=1 - \frac{1}{125}=\frac{124}{125} \). \( P(B'|A')=1 - P(B|A')=1 - \frac{70}{992}=\frac{992 - 70}{992}=\frac{922}{992}=\frac{461}{496} \). Then \( P(A' \cap B')=P(B'|A') \cdot P(A')=\frac{461}{496} \times \frac{124}{125}=\frac{461 \times 124}{496 \times 125} \). Simplify \( 124/496 = 1/4 \), so \( \frac{461}{4 \times 125}=\frac{461}{500} \). \( P(B')=1 - P(B)=1 - \frac{77}{1000}=\frac{923}{1000} \). Wait, no, better to use \( P(A'|B')=\frac{P(A' \cap B')}{P(B')} \). Alternatively, \( P(B')=1 - \frac{77}{1000}=\frac{923}{1000} \). \( P(A' \cap B')=P(B'|A') \cdot P(A')=(1 - \frac{70}{992}) \cdot \frac{124}{125}=\frac{922}{992} \cdot \frac{124}{125} \). Simplify \( 124/992 = 1/8 \), so \( \frac{922}{8 \times 125}=\frac{922}{1000}=\frac{461}{500} \). Then \( P(A'|B')=\frac{461/500}{923/1000}=\frac{461 \times 1000}{500 \times 923}=\frac{461 \times 2}{923}=\frac{922}{923} \) (wait, 922 and 923: 923=13×71, 922=2×461, so no simplification. Wait, maybe I made a mistake. Wait, \( P(B|A')=\frac{70}{992} \), so \( P(B'|A')=1 - \frac{70}{992}=\frac{922}{992} \). \( P(A')=\frac{124}{125} \), so \( P(A' \cap B')=\frac{922}{992} \times \frac{124}{125} \). 124/992 = 1/8, so \( \frac{922}{8 \times 125}=\frac{922}{1000}=\frac{461}{500} \). \( P(B')=1 - P(B)=1 - \frac{77}{1000}=\frac{923}{1000} \). So \( P(A'|B')=\frac{461/500}{923/1000}=\frac{461 \times 2}{923}=\frac{922}{923} \approx 0.9989 \).

Answers:

f) No
g) \( \frac{1}{11} \)
h) \( \frac{922}{923} \) (or simplified as above)

(Note: For f, the key is that \( P(A \cap B) eq P(A)P(B) \), so answer is no. For g, using \( P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{7/1000}{77/1000}=\frac{1}{11} \). For h, detailed calculation as above gives \( \frac{922}{923} \).)

Answer:

s:
f) No
g) \( \frac{1}{11} \)
h) \( \frac{922}{923} \) (or simplified as above)

(Note: For f, the key is that \( P(A \cap B)
eq P(A)P(B) \), so answer is no. For g, using \( P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{7/1000}{77/1000}=\frac{1}{11} \). For h, detailed calculation as above gives \( \frac{922}{923} \).)