Question

A certain disease occurs most frequently among older women. Of all age groups, women in their 60s have the highest rate of breast cancer. The National Cancer Institute (NCI) compiles U.S. epidemiology data for a number of different cancers. The NCI estimates that 3.12% of women in their 60s get breast cancer. Mammograms are X-ray images of the breast used to detect breast cancer. A mammogram can typically identify correctly 81% of cancer cases and 92% of cases without cancer. If a woman in her 60s gets a positive mammogram, what is the probability that she indeed has breast cancer?

Answer 1

Answer:

The probability that she indeed has breast cancer is 0.0276.

Step-by-step explanation:

We are given that a certain disease occurs most frequently among older women. Of all age groups, women in their 60's have the highest rate of breast cancer.

The NCI estimates that 3.12% of women in their 60's get breast cancer. A mammogram can typically identify correctly 81% of cancer cases and 92% of cases without cancer.

Let Probability that women in their 60's get breast cancer = P(BC) = 0.0312

Probability that women in their 60's does not get breast cancer = P(BC') = 1 - P(BC) = 1 - 0.0312 = 0.9688

Also, let P = event that mammograms correctly detect positive results for breast cancer

So, Probability that mammograms correctly detect positive results given that women actually has breast cancer = P(P/BC) = 0.81

Probability that mammograms detect correctly given that women actually does not has breast cancer = P(P/BC') = 0.92

Now, to find the probability that she indeed has breast cancer given the fact that woman in her 60's gets a positive mammogram, we will use Bayes' Theorem;

The Bayes' theorem is given by;

The Bayes' theorem states that the conditional probability of an event, say $A_k$ given that another event, say X has already occurred is given by:

         $P(A_{k}|X)=\frac{P(X|A_{k})P(A_{k})}{\sum\limits^{k}_{i=1}{P(X|A_{i})P(A_{i})}}$

Similarly,     P(BC/P) = $\frac{P(BC) \times P(P/BC)}{P(BC) \times P(P/BC) + P(BC') \times P(P/BC')}$

                 = $\frac{0.0312 \times 0.81}{0.0312 \times 0.81 + 0.9688 \times 0.92}$

                   = $\frac{0.025272}{0.916568}$ = 0.0276

Hence, the required probability is 0.0276.

Answer 2

Answer:

The probability that a woman in her 60s has breast cancer given that she gets a positive mammogram is 0.0276.

Step-by-step explanation:

Let a set be events that have occurred be denoted as:

S = {A₁, A₂, A₃,..., Aₙ}

The Bayes' theorem states that the conditional probability of an event, say Aₙ given that another event, say X has already occurred is given by:

$P(A_{n}|X)=\frac{P(X|A_{n})P(A_{n})}{\sum\limits^{n}_{i=1}{P(X|A_{i})P(A_{i})}}$

The disease Breast cancer is being studied among women of age 60s.

Denote the events as follows:

B = a women in their 60s has breast cancer

+ = the mammograms detects the breast cancer

The information provided is:

$P(B) = 0.0312\\P(+|B)=0.81\\P(+|B^{c})=0.92$

Compute the value of P (B|+) using the Bayes' theorem as follows:

$P(B|+)=\frac{P(+|B)P(B)}{P(+|B)P(B)+P(+|B^{c})P(B^{c})}$

            $=\frac{(0.81\times 0.0312)}{(0.81\times 0.0312)+(0.92\times (1-0.0312)}$

            $=\frac{0.025272}{0.025272+0.891296}$

            $=0.02757\\\approx0.0276$

Thus, the probability that a woman in her 60s has breast cancer given that she gets a positive mammogram is 0.0276.