RBSE Solutions for Class 11 Maths Chapter 16 Probability Miscellaneous Exercise

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 16 Probability Miscellaneous Exercise Textbook Exercise Questions and Answers.

RBSE Class 11 Maths Solutions Chapter 16 Probability Miscellaneous Exercise

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:
In box, No. of red (R) marbles =10
and No. of blue (B) marbles = 20
No. of green (G) marbles = 30
Thus, no. of total marbles in box = 10 + 20 + 30 = 60
Thus, no. of elements in sample space n(5) = 60

(i) No. of ways to select 5 blue marbles out of 20 blue (B) marbles = ²⁰C₅
Thus, total results to draw blue marbles n(B) = ²⁰C₅
and no. of ways to select 5 marbles out of 60 marbles
= ⁶⁰C₂
Thus, probability to select 5 blue marble
n(B) = ²⁰C₅
and no. of ways to select 5 marbles out of 60 marbles
= ⁶⁰C₅
Thus, Probability to select 5 blue marbles = \(\frac{{ }^{20} C_5}{{ }^{60} C_5}\)

(ii) No. of ways to select 5 green (G) marble = ³⁰C₅
Thus, n(G) = ³⁰C₅
and n(S) = ⁶⁰C₅
Thus, probability that all five marbles are of green (G) colour
P(G) = \(\frac{{ }^{30} C_5}{{ }^{60} C_5}\)
Then, Probability that all five marbles are not of green colour
P(G') = 1 - P(G) = 1 - \(\frac{{ }^{30} C_5}{{ }^{20} C_5}\)
Probability to select at least 1 green colour marble
= 1 - \(\frac{{ }^{30} C_5}{{ }^{60} C_5}\)

 

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:
Total no. of ways to draw 4 cards from a deck = ⁵²C₄
Thus, no. of elements in sample space n(S) = ⁵²C₄
Since in a deck of cards, there are 13 cards of spade and 13 of diamond.
Thus, total no of ways to select 3 diamond and 1 spade card.
= ¹³C₃ × ¹³C₁,
If event E : ‘Draw 3 diamond and 1 spade card’ ,Then, n(E) = nC₃ × 13C₁
Then, probability of event E
P(E) = \(\frac{n(E)}{n(S)}=\frac{{ }^{13} C_3 \times{ }^{13} C_1}{{ }^{52} C_4}\)
Thus, probability to draw 3 diamond and 1 spade card
= \(\frac{{ }^{13} C_3 \times{ }^{13} C_1}{{ }^{52} C_4}\)

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:
We have,
Two faces of die numbered with ‘1’.
Three faces of die each with numbered ‘2’.
And one face of die numbered with ‘3’.
Thus, in throwing a die, the sample space S = { 1, 1, 2, 2, 2, 3}
Then, n(S) = 6 (no. of sample points)
If event A : Occurrence of number ‘ 1 ’
Then, A = {1, 1}
and n(A) = 2
Event- B : Occurrence of number ‘2’
Then, B = {2, 2, 2}
and n(A) = 3
Event C : Occurrence of number ‘3’
Then, C = {3}
and n(C) = 1
P(2) = P(B) = \(\frac{n(B)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)
P(1) = P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)
and P(E) = P(C) = \(\frac{n(C)}{n(S)}=\frac{1}{6}\)
Thus, (i) Probability that 2 appear on dice
P(2) = \(\frac{1}{2}\)

(ii) Probability that 1 or 3 appear on dice P(1 or 3) = P(1) + P(3)
Since two events are Mutually exclusive
= \(\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}\)
Thus, P(1 or 3) = \(\frac{1}{2}\)

(iii)P(3-not) = 1 - P(3) = 1 - \(\frac{1}{6}=\frac{5}{6}\)

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:
We have,
No. of sold lottery tickets = 10,000
Thus, no. of elements in sample space n(S) = 10,000
Total no. of awards = 10

(a) If one ticket is buy then no. of ways to select 1 ticket out of 10 = ¹⁰C₂
and no. of ways to select 1 award out of 10,000 10,000
Thus, probability to not to get award
= \(\frac{{ }^{10} C_1}{{ }^{10,000} C_1}=\frac{10}{10,000}=\frac{1}{1,000}\)

(b) If 2 tickets are bought, then No. of ways to select 2 tickets out of 10 tickets = ¹⁰C₂
and total no. of ways to select 2 tickets out of 10,000
= ^10,000C₂

Then, probability P(2) to get 2 awards = \(\frac{{ }^{10} C_2}{{ }^{10,000} C_2}\)
Thus, probability to not get 2 awards

(c) When 10 tickets are bought, then no. of ways to t 10 awards out of 10 = ¹⁰C₁₀ = 1
and no. of ways to select 10 tickets out of 10,000 = ^10,000C₁₀

Thus, probability to get 10 awards P(10) = \(\frac{{ }^{10} C_{10}}{{ }^{10,000} C_{10}}\)
Then, probability to not get 10 awards = 1 - P(10)
= 1 - \(\frac{1}{{ }^{10,000} C_{10}}\)

Alternative Method :
No. of sold lottery tickets = 10,000 .
and total no. of awards =10
Then no. of tickets on which no award occur = 10,000 - 10 = 9,990
(a) Thus, buying 1 ticket, probability to not get award
= \(\frac{{ }^{9,990} C_1}{{ }^{10,000} C_1}=\frac{9,990}{10,000}=\frac{999}{1,000}\)

(b) Now , buying 2 tickets, probability to not get award =\(\frac{{ }^{9,990} C_2}{{ }^{10,000} C_2}\)

(c) Now, buying 10 tickets, probabilitys to not get award =\(\frac{{ }^{9,990} C_{10}}{{ }^{10,000} C_{10}}\)

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:
Let E: class of 40 students
and F : class of 60 students
Thus, number of elements in sample space .
n(S) = 100
Again, out of 100 students, 2 students can be arranged in classes in the following ways = ¹⁰⁰C₂ = \(\frac{100 !}{2 !(100-2) !}\)
= \(\frac{100 \times 99 \times(98 !)}{2 \times 1 \times(98 !)}\) = 50

(a) Again, no. of ways, both the students enter in section E
= ⁴⁰C₂ = \(\frac{40 !}{2 !(40-2) !}\)
= \(\frac{40 \times 39 \times(38 !)}{2 \times 1 \times(38 !)}\) = 780

and no. of ways, both the students enter in section F
= ⁶⁰C₂ = \(\frac{60 !}{2 !(60-2) !}\)
= \(\frac{60 \times 59 \times(58 !)}{2 \times 1 \times(58 !)}\) = 1770

Then, probability that both the students enter in section E
P(E) = \(\frac{{ }^{40} C_2}{{ }^{100} C_2}=\frac{780}{4950}=\frac{26}{165}\)
and probability that both the students enter in section F
P(E) = \(\frac{{ }^{60} C_2}{{ }^{100} C_2}=\frac{1770}{4950}=\frac{59}{165}\)

We have, both the students will not be enter in section E and F together. Thus, event E and F are mutually exclusive , i.e., E ∩ F = Φ
and .P(E ∩ F) = Φ = 0
Probability that both the students will enter in section E or F will be P(E ∪ F).
Thus, P(E ∪ F) = P(E) + P(F)
= \(\frac{26}{165}+\frac{59}{165}\)
= \(\frac{85}{165}=\frac{17}{33}\)
Thus, probability that you and your friend be in same section = \(\frac{17}{33}\).

(b) When you and your friend are in different sections then total no. of ways of section
⁴⁰C₁ × ⁶⁰C₁ = 40 × 60 = 2400
Then, probability that you and your friend enter in different sections = \(\frac{2400}{4950}=\frac{16}{33}\)
Thus, required probability = \(\frac{16}{33}\)

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:
We have, 3 letters are send to 3 persons in 3 envelopes with addressed.
Thus, no. of ways to arrange 3 letters in 3 different -envelopes = 3! = 3 × 2 × 1 = 6
Again, if two lettters are inserted in wrong envelopes then third letter will be inserted in wrong envelope.
Thus, no. of ways not to inserting any letter in correct envelope = 2! = 2 × 1 = 2
Thus, probability that all letters are not inserted in correct envelopes
P (not correct address) = \(\frac{2}{6}=\frac{1}{3}\)
Then probability that all letters are inserted in correct envelopes
P(correct address) = 1 - P (not correct address)
= 1 - \(\frac{1}{3}=\frac{2}{3}\)
Thus, probability that at least one letter is in its proper envelope = \(\frac{2}{3}\)

Question 7.
A and B are two events such that P(A) = 0 - 54, P(B) = 6 69mdP(A ∩ B) = 0.35.Find:
(i)P(A ∪ B),
(ii)P(A' ∩ B),
(iii) P(A ∩ B')
(iv)P(B ∩ A')
Answer:
We have, for two events A and B
P(A) = 0.54,
P(B) = 0.69
and P(A ∩ 8) = 0.35
(i) P(A ∪ B)=P(A) + P(B) - P(A ∩ B)
= 0.54 + 0.69 - 0.35 = 0.88
Thus, P(A ∪ B) = 0. 88

(ii) P(A' ∩ B') = P(A ∪ B)' [By De-Morgan’s law]
= 1 - P(A ∪ B)
= 1 - 0.88 = 0.12
Thus,P(A' ∩ B')= 0.12

(iii )P(A ∩ B') = P(A) - P(A ∩ B)
= 0.54 - 0.35 = 0.19
Thus, P(A ∩ B') = 0.19

(iv) P(B ∩ A') = P(B) - P(A ∩ B)
= 0.69 - 0.35 = 0.34
Thus, P(B ∩ A') = 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 :

Name	Sex	Age in years
1. Harish	M	30
2. Rohan	M	33
3. Sheetal	F	46
4. Alis	F	28
5. Salim	M	41

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:
We have,
No. of persons in the managing committee = 5
No. of male workers = 3
No. of workers of age more than 35 years = 2
Now, no. of ways to select a member as a spokesman out of 5 members ⁵C₁ =5
then n(S) = 5
and no. of ways to select 1 male member out of 3 male for spokesperson = ³C₁ =3
⇒ n(M) = 3
Thus, probability that selected spokesperson be male
P(Male) = P(M) = \(\frac{{ }^3 C_1}{{ }^5 C_1}=\frac{n(M)}{n(S)}=\frac{3}{5}\)
Again, since two persons are of age more than 35 years.
n(age more than 35 years) = 2
Thus, probability that selected person is of the age more than 35 years
n(more than 35 years) = \(\frac{n(\text { more than } 35 \text { years })}{n(S)}=\frac{2}{5}\)
Again, no. of persons which are male as well as of age more than 35 years = 1
Thus, probability that selected person is male as well as of age more than 35 years P (male and above 35 years)
= P(M ∩ above 35 years) = \(\frac{1}{5}\)
Then, probability that selected person is male or above 35 years
= P(M or above 35 years)
=P(M ∪ above 35 years)
Thus, P(M ∪ above 35 years)
= P(M) + P (above 35 years) - P(M ∩ above 35 years)
= \(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}=\frac{4}{5}\)
Thus, required probability = \(\frac{4}{5}\)

Question 9.
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 ?
(ii) the repetition of digits is not allowed ?
Answer:
Given digits are 0, 1, 3, 5 and 7.
Now, four digit numbers will be greater than 5000 when 5 or 7 be placed at thousand’s place.
Thus, no. of ways to select 5 or 7 at thousands place
= ²P₂
= \(\frac{2 !}{(2-2) !}=\frac{2 \times 1}{0 !}\)
= 2 × 1 = 2
Now, after selecting 5 or 7, four digits are left.
They will be either 0, 1, 3, 5 or 0, 1,3, 7.

Now, no. of ways to fill three places (unit, tens, hundred) by four digits = ⁴P₃
= \(\frac{4 !}{(4-3) !}\)
= \(\frac{4 \times 3 \times 2 \times 1}{1}\) = 24

Thus, total numbers greater than 5000 formed . bv given digits = ²P₂ × ⁴P₃ = 2 × 24 = 48
Numbers so formed will be divisible by 5 only when 0 or 5 is placed at unit place.
Thus, no. of ways to select 0 or 5 at unit place = ²P₂ = 2!
and no. of ways to select three digits at remaining three places (tens, hundred, thousand) = 3!
Thus, no. of four digit numbers divisible by 5
= 3! x 2! + 3!
= 3 × 2 × 1 × 2 × 1 + 3 × 2 × 1
= 12 + 6
= 18
Thus, when digits are not repeated then probability to get four digit numbers divisible by 5 = \(\frac{18}{48}=\frac{3}{8}\)
(ii) If digits are repeated then 4 digit numbers greatei than 5000 = 2! × (5)³
=250 [Since, out of 5 digits any digit can be repeated 3 times]
Then, no. of numbers divisible by 5
= 2! × 5² × 2! = 100
Thus, when digits are repeated, probability to get foa; digit no. divisible by 5
= \(\frac{100}{250}=\frac{2}{5}\)

Question 10.
The number lock of a suitcase has 4 wheels, each labelled with ten digits Le., 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 ?
Answer:
We have,
There are 4 wheels in a lock and each wheel is numbered 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
To open the lock, a sequence of four digit is required.
Then no. of ways to select 4 digits out of 10 in sequence = ¹⁰P₄
= \(\frac{10 !}{(10-4) !}=\frac{10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\) = 5040

Since lock open with only a specify sequence.
Thus, required probability = \(\frac{1}{5040}\)