Q.3.10 Two percent of women age 45 who ... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 3: Q.3.10 (page 107)

Two percent of women age $45$ who participate in routine screening have breast cancer. Ninety percent of those with breast cancer have positive mammographies. Eight percent of the women who do not have breast cancer will also have positive mammographies. Given that a woman has positive mammography, what is the probability she has breast cancer?

Short Answer

Expert verified

The probability that the woman has breast cancer is $0.1867$

Step by step solution

Step 1:Given Information

Given that two percent of women age $45$ who participate in routine screening have breast cancer. Ninety percent of those with breast cancer have positive mammographies. Eight percent of the women who do not have breast cancer will also have positive mammographies.

Explanation

$B$ =Breast cancer

$C$ =Positive mammographies

$P (B) = 2 % = 0.02$

$P (C 鈭� B) = 90 % = 0.9$

$P (C 鈭� B^{c}) = 8 % = 0.08$

Use the complement rule:

$P (A^{c}) = P (not 鈦� A) = 1 鈭� P (A)$

$P (B^{c}) = 1 鈭� P (B) = 1 鈭� 0.02 = 0.98$

Explanation of Bayes's Theorem

Use the equation

P (A_{i} 鈭� B) = \frac{P (B 鈭� A_{i}) P (A_{i})}{{鈭�}_{j = 1}^{k} 鈥� P (B 鈭� A_{j}) P (A_{j})}

$P (B 鈭� C) = \frac{P (C 鈭� B) P (B)}{P (C 鈭� B) P (B) + P (C 鈭� B^{c}) P (B^{c})}$

Substitute the value,

$= \frac{0.9 脳 0.02}{0.9 脳 0.02 + 0.08 脳 0.98}$

$= \frac{0.018}{0.018 + 0.0784}$

$= \frac{0.018}{0.0964}$
$= \frac{180}{964}$

We get,

$= \frac{45}{241}$

localid="1648546616788" $= 0.1867$

Final Answer

The probability she has breast cancer is $\frac{45}{241} = 0.1867$ .

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Day	P(mail/accepted)	P(mail/rejected)
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91影视

Short Answer

Step by step solution

Step 1:Given Information

Explanation

Explanation of Bayes's Theorem

Final Answer

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