How can we solve this using a Venn diagaram?
12 out of 1000 women at age forty who participate in routine screening have breast cancer. 880 out of 1000 women with breast cancer will get positive mammographies. 90 out of 1000 women without breast cancer will also get positive mammographies. If 100 women in this age group undergo a routine screening, about what fraction of women with positive mammographies will actually have breast cancer?
To solve this problem using a Venn diagram, we can visualize the information given and use it to calculate the fraction of women with positive mammographies who actually have breast cancer.
Step 1: Draw two intersecting circles to represent the sets of women with and without breast cancer. Label one circle as "Breast Cancer" and the other as "No Breast Cancer."
Step 2: Determine the size of each set based on the given percentages. Since 12 out of 1000 women at age forty have breast cancer, we can represent this by shading a small portion (12/1000) of the "Breast Cancer" circle. For the "No Breast Cancer" circle, since 90 out of 1000 women without breast cancer will have a positive mammography, we shade a small portion (90/1000) of this circle.
Step 3: Calculate the overlap between the two circles. We know that 880 out of 1000 women with breast cancer will have positive mammographies. So, shade a larger portion (880/1000) of the "Breast Cancer" circle to indicate women with both breast cancer and a positive mammography.
Step 4: Now, we can find the fraction of women with positive mammographies who actually have breast cancer by comparing the shaded areas. To do this, divide the area of the overlap (shared between the two circles) by the total shaded area in the "Positive Mammography" region.
Step 5: Once you have calculated this fraction using the Venn diagram, you will have the answer to the question: about what fraction of women with positive mammographies will actually have breast cancer?
Keep in mind that you may need to use proportional reasoning or calculate using exact values if the numbers are not easily divisible.