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### Math Assignment Class XI Ch -16 | Probability

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# **Math Assignment Class XI Ch -16 | Probability**

**Extra questions of chapter 16 probability class 11 with answers and hints to the difficult questions. strictly according to the cbse syllabus. Important and useful math. assignment for the students of class 11**

Given P(A) =3/5 and PB) =1/5 . If A and B are mutually
exclusive events, then find P (A or B)

Ans: 4/5

Question 2:

Probability of not getting 12 as the sum in a throw of two
dice is

Ans: 35/36

Question 3:

A card is drawn from a deck of 52 cards. Find the
probability of getting a king or a heart or a red card.

Ans: 7/13

Question 4:

Two balls are drawn simultaneously from a bag containing 5
white and 7 red balls, find the probability that both the balls are red.

Ans 7/22

Question 5:

Define mutually exclusive events.

Ans: A & B are said to be mutually exclusive
events if A ⋂ B = Ñ„ or There is no common element in A and B.

Question 6:

A
coin is tossed and then a dice thrown write the sample space hence find the probability of getting a head and a prime number

Ans: 3/12 or 1/4

Question 7:

The probability of two events A and B are 0.25 and 0.50 respectively. The probability of their simultaneous occurrence (occurrence of both) is 0.14. Find the probability that neither A nor B occurs.

Ans: 0.39

Question 8:

The probability of happening of an event A is 0.5 and that
of B is 0.3. If A and B are mutually exclusive events then find the probability
of neither A nor B.

Ans 0.2

Question 9:

In a job interview for 4 posts, 5 boys and 3 girls
appeared. If selection of each candidate is equiprobable then find the
probability that

i) 3 boys and 1 girl or 1 boy and 3 girls are selected.

Ans 1/2

(ii) At most one girl is selected

Ans 1/2

Question 10:

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

(i) all Kings Ans: 1/7735

(ii) 3 Kings Ans: 9/1547

(iii) at least 3 Kings. Ans 46/7735

Question 11:

If an integer from 1 through 1000 is chosen at random, then
find the probability that the integer is a multiple of 2 or a multiple of 9.

Solution.

Multiple of 2 from 1 to 1000 are 2, 4, 6, 8,
..., 1000

Let n be the number of terms of above series.

∴ nth term = 1000

2 + (n - 1)2 = 1000

2 + 2n -
2 = 1000

2n = 1000

∴ n = 500

The number of multiple of 2 are 500.

The multiple of 9 are 9, 18, 27, ..., 999

Let m be the number of term in above series.

∴ m^{th} term = 999

9 + (m - 1)9 = 999

9 + 9m
– 9 = 999

9m = 999

∴ m = 111

The number of multiple of 9 are 111.

The multiple of 2 and 9 both are 18, 36, ..., 990

Let p be the number of terms in above series.

∴ pth term = 990

18 + (p - 1)18 = 990

18 + 18p – 18 = 990

18p = 990

P = 55

The number of multiple of 2 and 9 are 55.

∴ Number of multiple of 2 or 9 = 500 + 111 − 55 = 556

∴ Required probability = 556/1000 = 0.556

Question 12:

If A and B are mutually exclusive events, P (A) = 0 35. and
P (B) = 0.45, then Find

(i)
P(A’) Ans
0.65

(ii)
P(B’) Ans: 0.55

(iii)
P(A⋃B) Ans:
0.80

(iv)
P(A ⋂B) Ans: 0

(v)
P(A⋂B’) Ans 0.35

(vi)
P(A’⋂B’) Ans:
0.2

Question 13

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that

(i) all the three balls are white.

Ans: 5/143

(ii) all the three balls are red.

Ans: 28/143

(iii) one ball is red and two balls are white.

Ans: 40/143

Question 14

If a card is drawn from a deck of 52 cards, then find the probability of getting a king or a heart or a red card.

Ans: 28/52

Question 15

A drawer contains 50 bolts and 150 nuts. Half of the bolt and half of nuts
are rusted. If one of then is chosen at random, what is the probability that it
is rusted or bolt ?

Ans: 125/200 = 5/8

Question 16

A box contains 24 balls from 1 to 24. One ball is drawn at
random. Find the probability that ball drawn has a number which is a multiple
of 3 or 4.

Ans 1/2

Question 17

A, B, C are three mutually exclusive and exhaustive events
associated with a random experiment

Find P (A) given that P(B) = P(A) and P(C) = P(B)

Ans: 4/13

Solution Hint: Use P(A) + P(B) + P(C) = 1

Question 18

Two candidates
Sunil and Ravi appeared in an interview. The probability that Sunil will
qualify the interview is 0.04 and that Ravi will qualify the interview is 0.2.
The probability that both will qualify the interview is 0.03. Find the
probability that

(i) Both Sunil and Ravi will not qualify the interview.

Answer: 0.79

Solution Hint: P(A’ ⋂B’) = P(A ⋃B)’ = 1- P(A ⋃B)

(ii) Only one of them will qualify the interview.

Answer: 0.18

Solution Hint : P(A⋂B’) + P(A’⋂B) = P(A) – P(A⋂B) + P(B) – P(A⋂B)

Question 19

A and B are two mutually
exclusive and exhaustive events of a random experiment such that P(A) = 6[P(B)]^{2}
where P(A) and P(B) denotes probabilities of A and B respectively. Find P(A) and
P(B).

Ans: P(A) = 2/3, P(B) = 1/3

Solution Hint: P(A) + P(B) = 1

6[P(B)]^{2} + P(B) = 1

Now putting P(B) = x and then solve the quadratic equation for the value of x

Question 20

A box contains 10 bulbs, of which
three are defective. If a random sample of 5 bulbs is drawn, find the probabilities
that the sample contains

(i) Exactly two defective bulbs

Ans: 5/12

Solution Hint: Use

(ii) At the most one defective
bulb.

Ans: 1/2

Solution Hint: Use

Question 21

In
a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both
NCC and NSS. If one of these students is selected at random, find the probability
that

(i)
The student opted for NCC or NSS.

Ans:
19/30

(ii)
The student has opted neither NCC nor NSS.

Ans:
11/30

(iii)
The student has opted NSS but not NCC.

Ans: 2/15

Question 22

Out
of 100 students, two sections of 40 and 60 are formed. If you and your friend are
among the 100 students, what is the probability that

(a) you both enter the same section?

Ans:
17/33

(b) you both enter the different sections?

Ans: 16/33

Question 23

If
4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3,
5, and 7, what is the probability of forming a number divisible by 5 when,

(i) the digits are repeated?

Ans : 99/249

(ii) the repetition of digits is not allowed?

Ans: 18/48

Question 24

The probability that a student will pass the final examination in both English and Hindi is 0.5 and probability of passing neither is 0.1. If the probability of passing the English examination is 0.75.

What is the probability of passing the Hindi examination?

Ans: 0.65

Solution
Hint:

P(E

P(E'⋂H') = 0.1 ⇒ P(E⋃H)' = 0.1

⇒1- P(E⋃H) = 0.1

⇒P(E⋃H) = 0.9

Now using P(E⋃H) = P(E) + P(H) - P(E⋂H) we get

P(H) = 0.65

Question 25

An urn contains 7 white balls, 5 black balls and 3 red balls. Two balls are drawn at random. Find the probability that:

(a) both the balls are red

Ans: 1/35

Solution Hint : Using

(b) One ball is red and other is black

Ans: 1/7

Solution Hint : Using

Question 26

If the letters of the word ‘ALGORITHM’ are arranged at random in a row what is the probability the letters ‘GOR’ must remain together as a unit?

Solution.

Number of letters in the word ‘ALGORITHM’ = 9

If ‘GOR’ remain together, then considered it as 1 group.

∴ Number of letters = 6+1 = 7

Number of word, if ‘GOR’ remain together = 7!

Total number of words from the letters of the word ‘ALGORITHM’ = 9!

∴ Required probability = 7! / 9! = 1/72

Question 27

Two students A and B appeared in an examination. The probability that A will qualify the examination is 0.05 and that B will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02.

Based on the above information answer the following questions

a) What is the probability that both A and B will not qualify the examination ?

Ans: 0.87

Solution Hint

P(both will not qualify the examination) = P(A'⋂B')

P(A⋃B)' = 1- P(A⋃B) = 0.87

b) What is the probability that at least one of them will not qualify the examination?

Ans 0.98

Solution Hint

P(At least one of them will not qualify) = 1- P(both will qualify)

= 1- P(A⋂B) = 1-0.02 = 0.98

c) What is the probability that only one of them will qualify the examination?

Ans: 0.11

Solution Hint

P(Only one will qualify) = P(A - B) + P(B - A)

= P(A⋂B') + P(B⋂A')

= P(A) - P(A⋂B) + P(B) - P(A⋂B)

= 0.05 - 0.02 + 0.1 - 0.02 = 0.11

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