# Important Questions for Class 12 Chapter 13 - Probability

CBSE Board has laid down specific guidelines for the pattern of questions that a student of class 12th should be acquainted with. Adhering to these guidelines, we have come up with class 12th important questions which carries questions related to all topics in Mathematics.

**Important questions based on NCERT syllabus for Class 12 Chapter 13 - Probability:**

*Question 1*: Consider another example where a pack contains 5 blue, 3 red and 4 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 blue pens and 1 black pen?

*Solution*: Here, total number of pens = 12

Probability of drawing 1 blue pen = 5/12

Probability of drawing another blue pen = 5/12

Probability of drawing 1 black pen = 4/12

Probability of drawing 2 blue pens and 1 black pen = 5/12 * 5/12 * 4/12 = 25/288

*Question 2*: What is the probability of the occurrence of a number that is odd or less than 10 when a fair die is rolled.

*Solution*: Let the event of the occurrence of a number that is odd be ‘A’ and the event of the occurrence of a number that is less than 10 be ‘B’. We need to find P(A or B).

P(A) = 3/6 (odd numbers = 1,3 and 5)

P(B) = 9/6 (numbers less than 5 = 1,2,3 and 4)

P(A and B) = 5/6 (numbers that are both odd and less than 10 = 1,3, 5, 7 and 9)

Now, P(A or B) = P(A) + P(B) – P(A and B)

= 3/6 + 9/6 – 5/6

P(A or B) = 7/6.

*Question 3*: When two dice are rolled, find the probability of getting a greater number on the first die than the the second, given that the sum should equal 8.

*Solution*: Let the event of getting a greater number on the first die be G.

There are 3 ways to get a sum of 10 when two dice are rolled = {(5,5)(4,6)(6,4)}.

And there is only one ways where the number on the first die is greater than the one on the second given that the sum should equal 8, G = {(6,2)}.

Therefore, P(Sum equals 8) = 3/36 and P(G) = 1/36.

Now, P(G|sum equals 8) = P(G and sum equals 8)/P(sum equals 8)

= (3/36)/(1/36)

= 3