Important Questions for Class 11 Maths Chapter 16 - Probability

Important questions for class 11 Maths Chapter 16 – Probability are given here, which are taken from the previous year question paper. Go through all the problems provided here, which will help you to score good marks in class 11 final examination. Before solving the problems, be thorough with the concepts covered in the chapter – Probability. All the important questions in chapter 16 probability are solved here in a step by step procedure. Student can prepare these questions for the upcoming examinations because most of the questions are asked from the previous year question papers. Also, get important questions for class 11 Maths for all the chapters at BYJU’S.

Class 11 Maths Chapter 16 – Probability covers important concepts, such as random experiment, algebra of events, events and types of events, axiomatic approach to the probability

Also, check:

Class 11 Chapter 16 – Probability Important Questions with Solutions

The most important questions for class 11 Maths Chapter 16 probability are provided with solutions. It includes MCQs, short type and long type answers. Practice these problems in a smart way a achieve good marks in the examination.

Question 1:

If P(A) is ⅗. Find P (not A)

Solution:

Given that: P(A) = ⅗

To find P(not A) = 1 – P(A)

P (not A) = 1- ⅗

= (5-3)/5

= ⅖

Therefore, P(not A) = ⅖.

Question 2:

Find the probability that when a hand of 7 cards are drawn from the well-shuffled deck of 52 cards, it contains

(i) all kings (ii) 3 kings

Solution:

(i) To find the probability that all the cards are kings:

If 7 cards are chosen from the pack of 52 cards

Then the total number of combinations possible are: 52C7

= 52!/[7! (52-7)!]

= 52!/ (7! 45!)

Assume that A be the event that all the kings are selected

We know that there are only 4 kings in the pack of 52 cards

Thus, if 7 cards are chosen, 4 kings are chosen out of 4, and 3 should be chosen form the 48 remaining cards.

Therefore, the total number of combinations is:

n(A) = 4C4 x48C3

= [4!/4!0! ] x [48!/3!(48-3)!]

= 1 x [48!/3! 45!]

= 48!/3! 45!

Therefore, P(A) = n(A)/n(S)

= [48!/3! 45!] ÷[52!/ (7! 45!]

= [48! x 7!] ÷ [3!x 52!]

= 1/7735

Therefore, the probability of getting all the 7 cards are kings is 1/7735

(ii) To find the probability that 3 cards are kings:

Assume that B be the event that 3 kings are selected.

Thus, if 7 cards are chosen, 3 kings are chosen out of 4, and 4 cards should be chosen form the 48 remaining cards.

Therefore, the total number of combinations is:

n(B) = 4C3 x48C4

= [4!/3!(4-3)! ] x [48!/4!(48-4)!]

= 4 x [48!/4! 44!]

= 48!/3! 45!

Therefore, P(B) = n(B)/n(S)

= [4 x48!/4! 44!] ÷[52!/ (7! 45!]

= 9/1547

Therefore, the probability of getting 3 kings is 9/1547

Question 3:

An urn contains 6 balls of which two are red and four are black. Two balls

are drawn at random. The probability that they are of different colours is

(i)⅖ (ii)1/15 (iii)8/15 (iv)4/15

Solution:

A correct answer is an option (c)

Explanation:

Given that, the total number of balls = 6 balls

Let A and B be the red and black balls respectively,

The probability that two balls drawn are different = P(the first ball drawn is red)(the second ball drawn is black)+ P(the first ball drawn is black)P(the second ball drawn is red)

= (2/6)(4/5) + (4/6)(2/5)

=(8/30)+ (8/30)

= 16/30

= 8/15

Question 4:

A couple has 2 children. Find the probability that

(i) one of them is a boy

(ii) the older child is a boy

Solution:

Given that, a couple has 2 children,

Let B be a boy and G be a girl

Then the sample space, S = (BB, BG, GB, GG}

(i) The probability that one of them is a boy:

Let, A = atleast one of them is a boy {BB, BG, GB}

B = both are boys {BB}

Therefore, P(B/A) = P(A∩B)/ P(A)

= (¼)/(¾)

= ⅓

Therefore, the probability that one of them is a boy = ⅓

(ii) The probability that an older child is a boy:

Let, A = elder one is a boy {BB, BG}

B = both are boys {BB}

Therefore, P(B/A) = P(A∩B)/ P(A)

= (¼)/(2/4)

= ½

Therefore, the probability that elder one is a boy = 1/2

Question 5:

One card is drawn from a well-shuffled pack of 52 cards. What is the probability that a card will be

(i) a diamond (ii) Not an ace (iii)a black card (iv) not a diamond

Solution:

(i) the probability that a card is a diamond

We know that there are 13 diamond cards in a deck. Therefore, the required probability is:

P( getting a diamond card) = 13/52 = ¼

(ii) the probability that a card is not an ace

We know that there are 4 ace cards in a deck.

Therefore, the required probability is:

P(not getting an ace card) = 1-( 4/52)

= 1- (1/13)

=(13-1)/13

= 12/13

(iii) the probability that a card is a black card

We know that there are26 black cards in a deck.

Therefore, the required probability is:

P(getting a black card) = 26/52 = ½

(iv) the probability that a card is not a diamond

We know that there are 13 diamond cards in a deck.

We know that the probability of getting a diamond card is 1/4

Therefore, the required probability is:

P( not getting a diamond card) = 1- (1/4)

= (4-1)/4

= 3/4

Practice Problems for Class 11 Maths Chapter 16 Probability

Practice the problems given below:

  1. List the sample space is throwing a die?
  2. List the sample space in tossing a coin?
  3. List the space in the simultaneous toss of a die and a coin?
  4. In a single throw of two dice, find the probability of obtaining a total of 8.
  5. A bag contains 9 red and 12 white balls. One ball is drawn at random. Find the probability that the ball is drawn is red.
  6. What is the probability that the ordinary year has 53 Sundays?
  7. A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of its being a spade or a king.
  8. A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card?
  9. A number is chosen from the number 1 to 100. Find the probability of its being divisible by 4 or 6.
  10. Two cards are drawn at random from a pack of 52 cards. What is the probability that both the drawn cards are aces?

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