Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

44. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4 she tosses a coin once and noted whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 and 4 with the die?

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

44. Let, $E_{1} :$ event that outcome on the die is $5$ or $6$

$E_{2} :$ event that outcome on the die is $1, 2, 3$ or $4$

$A :$ event of getting exactly one head

$P (E_{1}) = \frac{2}{6} = \frac{1}{3}$

$P (E_{2}) = \frac{4}{6} = \frac{2}{3}$

$P (A / E_{1}) =$ probability of getting exactly one head by tossing the coin three times if she gets $5$ or $6$ $= \frac{3}{8}$

$P (A / E_{2}) =$ probabilit

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

43. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability if 0.3, if the second group wins. Find the probability that the new product introduced was by the second group.

20 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

43. Let, $E_{1} :$ event that first group will win

$E_{2} :$ event that second group will win

$A :$ event that new product will produce

$P (E_{1}) = 0.6 = \frac{6}{10} = \frac{3}{5}$

$P (E_{2}) = 0.4 = \frac{4}{10} = \frac{2}{5}$

$P (A / E_{1}) = P$ (introducing new product by group $A$ ) $= 0.7 = \frac{7}{10}$

$P (A / E_{2}) = P$ ( introducing new product by group $B$ ) $= 0.3 = \frac{3}{10}$

Therefore, by Baye’s theorem,

$P (E_{2} / A) =$ probability that n

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

42. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

42. Let, $E_{1} :$ event items produced by $A$

$E_{2} :$ event items produced by $B$

$X :$ event that produced item was found to be defective

$P (E_{1}) = 60 % = \frac{3}{5}$

$P (E_{2}) = 40 % = \frac{2}{5}$

$P (X / E_{1}) = P$ (items produced by machine $A$ which is defective) $= 2 % = \frac{2}{100}$

$P (X / E_{2}) = P$ ( items produced by machine $B$ which is defective) $= 1 % = \frac{1}{100}$

Therefore, by Baye’s

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

41. An insurance company insured 2000 scooter driver, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

41. Let, $E_{1} :$ event that the driver is scooter driver

$E_{2} :$ event that the driver is car driver

$E_{3} :$ event that the driver is truck driver

$A :$ event the person meets with accident

Total number of drivers $= 2000 + 4000 + 6000 = 12000$

$P (E_{1}) = P$ (driver is a scooter driver) $= \frac{2000}{12000} = \frac{1}{6}$

$P (E_{2}) = P$ (driver is a car driver) $= \frac{4000}{12000} = \frac{1}{3}$

$P (E_{3}) = P$ (dri

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

40. There are three coins. One is a two headed coin, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows head, what is the probability that it was the two headed coin?

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

40. Let, $E_{1} :$ event of choosing $Q$ headed coin

$E_{2} :$ event of choosing biased coin

$E_{3} :$ event of choosing unbiased coin

$P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$

Let $A$ be the event that shows head,

$P (A / E_{1}) = P$ (coin shows heads, given that it is a headed coin) $= 1$

$P (A / E_{2}) = P$ (coin showing up head, given that it is biased coin

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New answer posted•a year ago

Ncert Solutions 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?

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year 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 positiv

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

38. In answering a question on a multiple choice test a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses the answer, will be correct with probability 1/4. What is the probability that a student knows the answer given that he answered it correctly?

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

38. Let, $E_{1}$ :event that knows the answer

$E_{2}$ :event that he guesses the answer

$P (E_{1}) = \frac{3}{4}$ and $P (E_{2}) = \frac{1}{4}$

Let $A$ be the event that the answer is correct

Also, $P (A / E_{1}) = P$ (correct answer given that he knows) $= 1$

$P (A / E_{2}) = P$ (correct answer given that he guess) $= \frac{1}{4}$

Now, probability that he knows the an

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

37. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hosteller?

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

37. Let, $E_{1}$ :event that the student in hostel

$E_{2}$ : event that the student is day scholar

$A$ : be the event of the chosen student get grade $A$

$P (E_{1}) = 60 % = \frac{60}{100} = 0.6$

$P (E_{2}) = 40 % = \frac{40}{100} = 0.4$

$P (A / E_{1}) = P$ (student getting an $A$ grade is hostler) $= 30 %$ $= \frac{30}{100} = 0.3$

$P (A / E_{2}) = P$ ( student getting an $A$ grade is a day scholar) $= 20 % = \frac{20}{100} = 0.2$

Then, by Ba

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

36. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

36. Let $E_{1}$ be the event of selecting first bag

$E_{2}$ be the event of selecting second bag

$∴ P (E_{1}) = P (E_{2}) = \frac{1}{2}$

Let $A$ be the event of getting a red ball

$P (A / E_{1}) = P$ (drawing a red ball from first bag)

$= \frac{4}{8} = \frac{1}{2}$

$P (A / E_{2}) = P$ (drawing a red ball from second bag)

$= \frac{2}{8} = \frac{1}{4}$

Then, probability of drawing a ball from the first bag which is red, is

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New answer posted•a year ago

Ncert Solutions Maths class 12th Probability

35. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

35. Number of balls contain in win $= 5$ red

$= 5$ black

Let the red ball be drawn in the first attempt

$∴$ $P$ (drawing a red ball) $= \frac{5}{10} = \frac{1}{2}$

By question, if added two red balls to the win, then $7$ red balls and $5$ black balls contain.

$P$ (drawing a red ball) $= \frac{7}{12}$

Let a black ball be

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Ncert Solutions Maths class 12th

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