NCERT Solutions Class 11 Mathematics Probability Exercise 16-Misc

Question 1:
A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that                                                                                           
  (i) all will be blue?                         (ii) at least one will be green?

Answer:
Total number of marbles = 10 + 20 + 30 = 60
Number of ways of drawing 5 marbles from 60 marbles = ⁶⁰C₅
(i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles.
5 blue marbles can be drawn from 20 blue marbles in ²⁰C₅ ways.
So, probability that all marbles will be blue = ²⁰C₅ / ⁶⁰C₅
(ii) Number of ways in which the drawn marble is not green = ^{(20 + 10)}C₅ = ³⁰C₅
So, probability that no marble is green = ³⁰C₅ / ⁶⁰C₅
Hence, probability that at least one marble is green = 1 – ³⁰C₅ / ⁶⁰C₅

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?

Answer:
Number of ways of drawing 4 cards from 52 cards = ⁵²C₄
In a deck of 52 cards, there are 13 diamonds and 13 spades.
Number of ways of drawing 3 diamonds and one spade = ¹³C₃ * ¹³C₁
Thus, the probability of obtaining 3 diamonds and one spade = (¹³C₃ * ¹³C₁)/ ⁵²C₄
 
 

Question 3:
A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine                                                                                           
 (i) P(2)                                     (ii) P(1 or 3)                                          (iii) P(not 3)

Answer:
Total number of faces = 6
(i) Number faces with number ‘2’ = 3
So, P(2) = 3/6 = 1/2
(ii)P (1 or 3) = P (not 2) = 1 − P (2) = 1 – 1/2 = 1/2
(iii) Number of faces with number ‘3’ = 1
So, P(3) = 1/6
Hence, P(not 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.

Answer:
Total number of tickets sold = 10000
Number prizes awarded = 10
(i) If we buy one ticket, then
P (getting a prize) = 10/10000 = 1/1000
So, P (not getting a prize) = 1 – 1/1000 = 999/1000
(ii) If we buy two tickets, then
Number of tickets not awarded = 10,000 − 10 = 9990
P (not getting a prize) = ⁹⁹⁹⁰C₂ / ¹⁰⁰⁰⁰C₂
(iii) If we buy 10 tickets, then
P(not getting a prize) = ⁹⁹⁹⁰C₁₀ / 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?

Answer:
My friend and I are among the 100 students.
Total number of ways of selecting 2 students out of 100 students = ¹⁰⁰C₂
(a)The two of us will enter the same section if both of us are among 40 students or among
60 students.
So, then number of ways in which both of us enter the same section = ⁴⁰C₂ + ⁶⁰C₂
Probability that both of us enter the same section
= (⁴⁰C₂ + ⁶⁰C₂)/ ¹⁰⁰C₂
= {(40 * 39)/(2 * 1) + (60 * 59)/(2 * 1)}/{ (100 * 99)/(2 * 1)}
= {(40 * 39) + (60 * 59)}/(100 * 99)    
= {(4 * 39) + (6 * 59)}/(10 * 99)
= {(4 * 13) + (2 * 59)}/(10 * 33)
= (52 + 118)/(10 * 33)
= 170/(10 * 33)
= 17/33
(b) P(we enter different sections) = 1 − P(we enter the same section)
                                                             = 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.

Answer:
Let L₁, L₂, L₃ be three letters and E₁, E₂, and E₃ be their corresponding envelops respectively.
There are 6 ways of inserting 3 letters in 3 envelops. These are as follows:
L₁ E₁, L₂ E₃, L₃ E₂
L₂ E₂, L₁ E₃, L₃ E₁
L₃ E₃, L₁ E₂, L₂ E₁
L₁ E₁, L₂ E₂, L₃ E₃
L₁ E₂, L₂ E₃, L₃ E₁
L₁ E₃, L₂ E₁, L₃ E₂
There are 4 ways in which at least one letter is inserted in a proper envelope.
Thus, the required probability = 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 ∩ B) = 0.35.
Find     (i) P(A ∪ B)                (ii) P(A´∩ B´)              (iii) P(A ∩ B´)                (iv) P(B ∩ A´)

Answer:
It is given that P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35
(i) We know that P (A ∪ B) = P(A) + P(B) − P(A ∩ B)
So, P(A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88
(ii) A′ ∩ B′ = (A ∪ B)′              [by De Morgan’s law]
So, P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12
(iii) P(A ∩ B′) = P(A) − P(A ∩ B) = 0.54 − 0.35 = 0.19
(iv) We know that:
       n(B ∩ A’) = n(B) – n(A ∩ B)
⇒ n(B ∩ A’)/n(s) = n(B) /n(s) – n(A ∩ B) /n(s)
So, P(B ∩ A’) = P(B) – P(A ∩ B)
                        = 0.69 – 0.35
                        = 0.34   

Question 8:
From the employees of a company, 5 persons are selected to represent them in the managing committee of the company.
Particulars of five persons are as follows:

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?

Answer:
Let E be the event in which the spokesperson will be a male and F be the event in which the
spokesperson will be over 35 years of age.
Accordingly, P(E) = 3/5 and P(F) = 2/5
Since there is only one male who is over 35 years of age,
P(E ∩ F) = 1/5
We know that:
P(E U F) = P(E) + P(F) – P(E ∩ F)
               = 3/5 + 2/5 – 1/5 = 4/5
Thus, the probability that the spokesperson will either be a male or over 35 years of age is 4/5.