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# Class 11 NCERT Solutions- Chapter 16 Probability – Miscellaneous Exercise on Chapter 16

• Last Updated : 28 Jan, 2021

### (i) all will be blue? (ii) at least one will be green?

Solution:

The total number of marbles in the box = 10 + 20 + 30 = 60 marbles

As 5 marbles are drawn from the box at a time, the number of ways drawing them = 60C5

i) all will be blue.

all the marbles would be blue when, out of 20 blue marbles, 5 are drawn.

So, the number of ways of drawing 5 blue marbles = 20C

Hence, the probability of drawing all the blue marbles =  20C5 Ã· 60C5

ii) at least one will be green.

the number of ways in which the drawn marble is not green = number of ways drawing red marble + number of ways drawing blue marbles

i.e 10C5 + 20C5 = (10+20)C5 = 30C5

the probability that no marble is green = 30C5

So, the probability of at least getting one green marble = 1 – the probability of no green marble =  1 – 30C5

### Question 2: 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

Solution:

the total number of ways in which 4 cards can be drawn from the deck = 52C4

In a deck of cards, there are 13 diamonds and 13 spades

number of ways of drawing 3 diamonds = 13C3

number of ways of drawing 1 spade = 13C1

total number of ways of drawing 3 diamonds and 1 spade = 13C3 x 13C1

Thus, the probability of obtaining 3 diamonds and 1 spade from a deck of cards = (13C3 x 13C1) Ã· 52C4

### Question 3: A die has two faces each with the number â€˜1â€™, three faces each with the number â€˜2â€™, and one face with the number â€˜3â€™. If the die is rolled once, determine

(I) P(2)

(ii) P(1 or 3)

(iii) P(not 3)

Solution:

The total number of faces in a dice = 6

(i) P(2)

the number of faces with the number ‘2’ = 3

so, the probability of getting 2 = P(2) = 3/6 = 1/2

(ii) P(1 or 3)

the number of faces with number ‘1’ = 2

the number of faces with number ‘3’= 1

total number of faces with either number ‘1’ or number ‘3’ = 1 + 2 = 3

so, the probability of getting either number ‘1’ or ‘3’ = P(1 or 3) = 3/6 = 1/2

Alternative Solution of (ii)

P (1 or 3 ) = P(not 2 ) = 1 – P(2) = 1-1/2 = 1/2

(iii) P(not 3 )

the number of faces with the number ‘3’ = 1

probability of getting 3 = P(3)= 1/6

so, the probability of not getting 3 = P(not 3) = 1 – P(3)

= 1-1/6 = 5/6

### Question 4: In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy

(a) one ticket

(b) two tickets

(c) 10 tickets

Solution:

the total number of lottery tickets sold = 10,000

number of prizes to be awarded = 10

(a) one ticket

probability of getting a prize = 10/10000 = 1/1000

probability of not getting a prize when one ticket is bought =

P(not getting a prize ) = 1 – P(getting a prize)

P(not getting a prize)  = 1 – 1/1000

= 999/1000

(b) two tickets

If we buy two tickets, then

the number of tickets not awarded = 10000-10

= 9990

number of ways in which we can buy two tickets that are not awarded = 9990C2

number of ways in which we can buy any two tickets = 10000C2

Hence, the probability of buying two non-prize tickets = (9990C2) Ã· (10000C2

(c) 10 tickets

If we buy two tickets, then

the number of tickets not awarded = 10000-10

= 9990

number of ways in which we can buy ten tickets that are not awarded= 9990C10

number of ways in which we can buy any ten tickets = 10000C10

Hence, the probability of buying ten non-prize tickets = (9990C10) Ã· (10000C10)

### Question 5: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?

(b) you both enter the different sections?

Solution:

Since, both I and my friend are among the 100 students,

so, the number of ways in which we can select 2 students among 100 students = 100C2

(a) you both enter the same section.

to be in the same section, we can either be in 40 students section or in 60 students section.

the total number of ways in which we both of us are in same section = 40C2 + 60C2

hence, the probability of both of us are in same section =

=  ( 40C2 + 60C2 ) Ã· (100C2)

= (( 39 x 40 ) + (59 x 60) ) Ã· (99 x 100)

= 17/33

(b) you both enter the different sections.

probability of entering the different sections = 1 – P(getting in the same section)

P(entering in different sections) = 1 – 17/33 = 16/33

### Question 6.Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

Solution :

Let the letters be denoted by L1, L2, L3.

And the envelopes be denoted by E1,E2,E3.

The letters are inserted into the envelopes and each envelope contains exactly one letter. The combinations that can be made are:

E1L1 E2L2 E3L3

E1L2 E2L1 E3L3

E1L2 E2L3 E3L1

E1L1 E2L3 E3L2

E1L3 E2L1 E3L2

E1L3 E2L2 E3L1

So, the total number of combinations that can be made is 6.

the combinations in which at least one letter is in its proper envelope: E1L1  E2L2  E3L3, E1L3  E2L2  E3L1, E1L1  E2L3  E3L2,            E1L2  E2L1  E3L3.

The number of such combinations is 4.

Hence, the probability of getting at least one letter in its proper envelope = 4/6 = 2/3

### Question 7: A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A âˆ© B) = 0.35. Find :

(i) P(A âˆª B)

Solution:

It is given that,  P(A) = 0.54, P(B) = 0.69 and P(A âˆ© B) = 0.35.

(i) P(A âˆª B)

we know that, P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

=> P(A âˆª B) = 0.54 + 0.69 – 0.35 = 0.88

we know that, by De Morgan’s Law, (AÂ´ âˆ© BÂ´) = (A âˆª B)Â´

=> P(AÂ´ âˆ© BÂ´) = P(A âˆª B)Â´

= P(AÂ´ âˆ© BÂ´) = 1 – P(A âˆª B)

= P(AÂ´ âˆ© BÂ´) = 1- 0.88 = 0.12

we know that, P(A âˆ© BÂ´) = P(A) – P(A âˆ© B)

=> P(A âˆ© BÂ´) = 0.54 – 0.35 = 0.19

we know that, n(B âˆ© AÂ´) = n(B) – n(A âˆ© B)

(n(B âˆ© AÂ´) Ã· n(S) ) = (n(B) Ã· n(S)) – (n(A âˆ© B) Ã· n(S))

=>  P(B âˆ© AÂ´) = 0.69 – 0.35 = 0.34

### A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

Solution:

Total number of employees in the company = 5

Number of males in employees = 3

number of persons above 35 years of age = 2

number of persons who are male and above 35 years of age = 1

the probability of selecting a male = 3/5 = P(E)

the probability of selecting a person above 35 years of age = 2/5 = P(F)

So, the probability of selecting the spokesperson who will be either a male or over 35 years = P(EâˆªF)

=> P(EâˆªF) P(E) + P(F) – P(Eâˆ©F)

=> P(EâˆªF) = 3/5 + 2/5 – 1/4 = 4/5

### (i) the digits are repeated? (ii) the repetition of digits is not allowed?

Solution:

(i) the digits are repeated.

Since we have to make 4-digit numbers greater than 5,000, so the left-most digit could either be 5 or 7.

the remaining 3 digits can be filled by any of the digits 0,1,3,5, and 7 as repetition is allowed. So, the total number formed greater than 5,000 are

= 2 x 5 x 5 x 5 = 250

A number is divisible by 5 if its one’s place is either filled by 0 or 5. the numbers that can be formed which are divisible by 5 are

= 2 x 5 x 5 x 2 = 100

So, the probability of forming numbers that are divisible by 5 =

=> 100/250 =  2/5

(ii) the repetition of digits is not allowed.

to make 4-digit numbers greater than 5,000 the left-most place, i.e, thousands place can be filled by either 5 or 7.

the remaining 3 places can be filled by the other 4 digits. So, the total numbers formed greater than 5,000 when repetition is not allowed are

= 2 x 4 x 3 x 2 = 48

A number is divisible by 5 if its one’s place is either filled by  5 or 0.

when the thousand’s place is taken by 5 and one’s place can only be filled with 0 to be divisible by 5. and the rest of the places can be filled by the remaining 3 digits.

the numbers that can be formed starting with 5 and ending at 0 are = 1 x 3 x 2x 1 = 6

and when the thousand’s place is taken by 7 and one’s place can be filled with 0 or 5 to be divisible by 5, the rest of the places can be filled by the remaining 2 digits.

the numbers that can be formed starting with 7 and ending at 5 or 0 are = 1 x 3 x 2 x 2 = 12

the total numbers formed greater than 5,000 and divisible by 5 when repetition of digits are not allowed = 12 + 6 = 18

So, the probability of forming numbers that are divisible by 5 when repetition is not allowed =

=> 18/48 = 3/8

### Question 10.The number lock of a suitcase has 4 wheels, each labeled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Solution:

Here, the number lock has 4 wheels each labeled with ten digits i.e from 0 to 9.

the number of ways of selecting 4 digits out of 10 digits = 10C4

as there are 4 wheels, each combination made can be arranged in 4! ways.

number of 4-digits combinations that can be made when no digit is repeated = 10C4 x 4! = 5040

As there is only one sequence which can open the suitcase.

Thus, the required probability is 1/5040.

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