Question

Probability trees 1. It is estimated that 100 out of 9,900 women who participated in a breast cancer screening had breast cancer. Of the women who participated in the screening and have breast cancer 80 out of 100 will have a positive mammogram. Of the women who participated in the screening and do not have breast cancer 950 out of 9800 will have a positive mammogram.

a) Create a probability tree

b) Calculate the probability that a woman has a positive mammogram given that she does not have breast cancer.

c) Calculate the probability that a woman has a negative mammogram.

d) Given that a woman has a negative mammogram calculate the probability that she has breast cancer.

e) Construct a contingency table comparing cancer diagnosis vs mammogram result.

f) Based on the contingency table what proportion of women have a negative mammogram result given that they have breast cancer.

g) Based on the contingency table what proportion of woman have breast cancer and a positive mammogram test.

h) Calculate the probability that a woman has breast cancer or a positive mammogram result.

i) Can the event “ Cancer Diagnosis or Positive test ” be considered independent events?

Answer 1

Solution-

Let the events that women having breast cancer diagnosis, breast cancer not diagnosis, positive mammogram test and negative memmogram test are denoted by C, C' , + ,and – respectively.

Now,

According to question, 100 out of 9,900 women who participated in a breast cancer screening had breast cancer.

So, P(C) = 100/9900 = 0.01 .....(1)

And number of women not having breast cancer is 9900-100 = 9800

So, P(C') = 9800/9900 = 0.99 ....(2)

The women who participated in the screening and have breast cancer 80 out of 100 will have a positive mammogram.

So, P(+/C)= 80/100 = 0.8 .....(3)

And women having breast cancer bit negative test are 100-80 = 20

So, P(-/C) = 20/100 = 0.2 ....4)

The women who participated in the screening and do not have breast cancer 950 out of 9800 will have a positive mammogram.

So, P(+/C') = 950/9800= 0.0969 ...(5)

Women not having breast cancer and negative mammogram test are = 9800-950 = 8850

So, P(-/C') = 8850/9800 = 0.9031 ....(6)

(a)

Uding the above information ,the probability tree of the above events is shown below-

(b)

The probability that a woman has a positive mammogram given that she does not have breast cancer.

P(+/C') = 0.0968 (from equation 5)

(c)

Total women = 9900

Women having negative test = 20 +8850=8870

So, the probability that a woman has a negative mammogram is

P(-)= 8870/9900 = 0.89

(d)

Number of women having negative memmogram test =20+8850 =8870

Number of women having breast cancer but negative test = 20

So, of it is given that a woman has a negative mammogram then the probability that she has breast cancer.

P(C/-) = 20/8870 = 0.00225479

(e)

The contingency table comparing cancer diagnosis vs mammogram result is shown below-

Number of women	having positive mammogram test	having negative memmogram test	Total
having breast Cancer diagnosis	80	20	100
having breast Cancer not diagnosis	950	8850	9800
Total	1030	8870	9900

(f)

Based on the contingency table the proportion of women have a negative mammogram result given that they have breast cancer is

= 20/100 = 0.2

(g)

Based on the contingency table the proportion of woman have breast cancer and a positive mammogram test is

=20/9900 = 0.00202

(h)

Number of women having breast cancer or positive test result = 9900 -8850 = 1050

The probability that a woman has breast cancer or a positive mammogram result.

P(C+) = 1050/9900 = 0.10606

(i)

No, the events “ Cancer Diagnosis or Positive test ” can be considered independent events because if a person has cancer diagnosis then he has more chances of positive test with respect to women who has not cancer diagnosis.So, two events are depend on each other.