Ask & Answer Question

Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

39. A laboratory blood test is 99% effective in detecting a certain disease when it is, in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

4 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

39. Let, $E_{1} :$ event person has disease

$E_{2} :$ event person has no disease

$A :$ be event that blood test is positive

As $E_{1}$ and $E_{2}$ are events which are complementary to each other,

Then, $P (E_{1}) + P (E_{2}) = 1$

$P (E_{2}) = 1 - P (E_{1})$

$P (E_{1}) = 0.1 % = \frac{0.1}{100} = 0.001$

$∴$ $P (E_{2}) = 1 - 0.001$

$= 0.999$

$P (A / E_{1}) = P$ (result is positive given that person has disease) $= 99 % = 0.99$

$P (A / E_{2}) = P$ (result is positive given that person has no disease) $= 0.5 % = 0.005$

Now, the probability that person has a disease, given that his test result is positive is $P (E_{1} / A)$

By using Baye’s theorem,

$P (E_{1} / A) = \frac{P (E_{1}) . P (A / E_{1})}{P (E_{1}) . P (A / E_{1}) + P (E_{2}) . P (A / E_{2})}$

$= \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005}$

$= \frac{0.00099}{0.00099 + 0.004995}$

$= \frac{0.00099}{0.005985} = \frac{990}{5985} = \frac{110}{665}$

$\Rightarrow P (E_{1} / A) = \frac{22}{133}$

0 Upvotes 0 Downvotes

Similar Questions for you

Let a die rolled till 2 is obtained. The probability that 2 obtained on even numbered toss is equal to

alok kumar singh

P (2 obtained on even numbered toss) = k (let)

P (2) = $\frac{1}{6}$

P ( $\bar{2}$ )= $\frac{5}{6}$

$k = \frac{5}{6} \times \frac{1}{6} + {(\frac{5}{6})}^{3} \times \frac{1}{6} + {(\frac{5}{6})}^{5} \times \frac{1}{6} + . . .$

$= \frac{\frac{5}{6} \times \frac{1}{6}}{1 - {(\frac{5}{6})}^{2}}$

$= \frac{5}{11}$

Given set S = {0, 1, 2, 3, ….., 10}. If a random ordered pair (x, y) of elements of S is chosen, then find probability that |x – y| > 5

alok kumar singh

If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability $= \frac{30}{11 * 11} = \frac{30}{121}$

A bag contains 8 balls (black and white). If four balls are chosen without replacement then 2W and 2B are found then the probability that number of white and black balls are same in bag is equal to

alok kumar singh

P (2W and 2B) = P (2B, 6W) × P (2W and 2B)

+ P (3B, 5W) × P (2W and 2B)

+ P (4B, 4W) × P (2W and 2B)

+ P (5B, 3W) × P (2W and 2B)

+ P (6B, 2W) × P (2W and 2B)

(15 + 30 + 36 + 30 + 15)

$= \frac{36}{126}$

$= \frac{18}{63}$

$= \frac{6}{21}$

$= \frac{2}{7}$

A coin is biased such that head has two chances than tails, what is the probability of getting 2 heads and 1 tail?

alok kumar singh

Let probability of tail is $\frac{1}{3}$

⇒ Probability of getting head = $\frac{2}{3}$

∴ Probability of getting 2 heads and 1 tail

$= (\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}) \times 3$

$= \frac{4}{27} \times 3$

$= \frac{4}{9}$

The coefficients a, b and c of the quadratic equation, ax² + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is:

Vishal Baghel

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l} a = 4 c = 1 \\ a = 2 c = 2 \\ a = 1 c = 4 \end{array}} 3 w a y s$

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability = $\frac{5}{216}$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question