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

13. An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple-choice questions and 400 difficult multiple choice questions. If a question is selected at random from the test bank, what is the probability that it will be an easy question given that it is a multiple-choice question?

2 Views | Posted 4 months ago

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

13. Here, there are two types of questions, True/False or Multiple-Choice Questions. They are divided into easy and difficult.

True/False: 300 easy, 200 difficult

$M C Q$ : 500 easy, 400 difficult

Let, $E =$ easy question

$M =$ multiple choice question

Total number of questions= $300 + 200 + 500 + 400 = 1400$

Total number of multiple-choice questions= $500 + 400 = 900$

Therefore, probability of selecting an easy $M C Q$ is

$P (E \cap M) = \frac{500}{1400} = \frac{5}{14}$

The probability of setting a $M C Q$ is

$P (M) = \frac{900}{1400} = \frac{9}{14}$

$∴$ $P (E / M) = \frac{P (E \cap M)}{P (M)} = \frac{\frac{5}{14}}{\frac{9}{14}} = \frac{5}{9}$

$∴$ Required probability is $\frac{5}{9}$

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