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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) atleast one will be green?

Solution

Total number of marbles = 10 + 20 + 30 = 60 Number of ways of drawing 5 marbles from 60 marbles = 60 C₅ .
(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 20 C₅ ways.
∴Probability that all marbles will be blue = \[\frac{{20{C_5}}}{{60{C_5}}}\] (ii) atleast one will be green.
There are 1 or 2 or 3 or 4 or 5 green marbles.
So Probability of atleast one green marble = 1 - P ( No green marbles)
there are 30 green out of 60 so there are 30 marbles which are not green.The selection of 5 marbles from it can be done in 30C₅ different ways.
So P ( no green marble ) = 30 C₅ / 60 C ₅
So Probability of atleast one green marble = 1 - P ( No green marbles)
\[ = 1 - \frac{{30{C_5}}}{{60{C_5}}}\]

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?

Solution

Number of ways of drawing 4 cards from 52 cards = n n= \[52{C_4}\] Let A be the event that 3 diamond cards and 1 spade card is drawn,
In a deck of 52 cards, there are 13 diamonds and 13 spades. Number of ways of drawing 3 diamond cards = 13 C ₃
Number of ways of drawing 1 spade card = 13 C ₁
Number of ways of drawing 3 diamond cards and 1 spade card = m
m = \[13{C_3} \times 13{C_1}\] \[P\left( A \right) = \frac{m}{n}\] \[ = \frac{{13{C_3} \times 13{C_1}}}{{52{C_4}}}\]

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)

Solution

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,
The sample space S will have 6 elements, as
S = { 1, 1, 2, 2, 2, 3 }
n = 6.
(i) P(2) = getting 2 on upper surface.
m = 3
So P(2) = m/n
= 3/6
= 1/2
(ii) P ( 1 or 3 ) = getting the number 1 or 3 on upper surface.
m = 3
So P(1 or 3) = m/n
= 3/6
= 1/2
(iii) 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

Total number of tickets sold = 10,000
Number prizes awarded = 10
The number of tickets not awared prizes = 10,000 - 10 = 9,990
(i) If we buy one ticket, then probability not getting prize , it means theticket is from 9,990 tickets.
P( not getting prize ) = m / n
= 9990 / 10000
= 0.999
(ii) If you have buy 2 tickets.
The selection of two tickets from 10,000 tickets is done by 10000 C₂ .
And 9990 tickets are not having prizes. We have to select 2 from it.
m = 9990 C₂ .
\[P\left( A \right) = \frac{m}{n}\] \[ = \frac{{9990{C_2}}}{{10000{C_2}}}\] (iii) If you have buy 10 tickets.
The selection of ten tickets from 10,000 tickets is done by 10000 C₁₀ .
And 9990 tickets are not having prizes. We have to select 10 from it.
m = 9990 C₁₀ .
\[P\left( A \right) = \frac{m}{n}\] \[ = \frac{{9990{C_{10}}}}{{10000{C_{10}}}}\]

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 sections? (b) you both enter the different sections?

Solution

My friend and I are among the 100 students.
Total number of ways of selecting 2 students out of 100 students = n = 100C ₂
(a) Let A be the event that both of us enter the same section
The two of us will enter the same section if both of us are among 40 students or among 60 students.
∴ Number of ways in which both of us enter the same section = m
m = \[40{C_2} + 60{C_2}\] ∴ Probability that both of us enter the same section
\[P\left( A \right) = \frac{m}{n}\] \[ = \frac{{40{C_2} + 60{C_2}}}{{100{C_2}}}\] \[ = \frac{{\frac{{40 \times 39}}{2} + \frac{{60 \times 59}}{2}}}{{\frac{{100 \times 99}}{2}}}\] \[ = \frac{{20\left( {2 \times 39 + 3 \times 59} \right)}}{{100 \times 99}}\] \[ = \frac{{3\left( {26 + 59} \right)}}{{5 \times 99}}\] \[ = \frac{{85}}{{5 \times 33}}\] \[ = \frac{{17}}{{33}}\] (b) you both enter the different sections
P ( you both enter the different sections)
= 1 - P( both enter the same sections )
= 1 - P ( A)
= 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

There are 3 letters to be inserted in 3 envelope . It can be done in 3! different ways.
So n = 3! = 6
Atleast one is in proper envelope, it means 1 will be proper envelope or 2 or 3
But when we insert 2 in proper atomatically third is in proper envelope only.
Let us consider 1 letter in proper envelope, it can be any one out of 3, there are 3 options.
All 3 letters are in proper envelope is posiible 1 times.
So for atleast 1 in proper envelope number of possible way = 3 + 1
So m = 4
\[P\left( A \right) = \frac{m}{n}\] \[ = \frac{4}{6}\] \[ = \frac{2}{3}\]

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 U B) (ii) P(A′ ∩ B′) (iii) P(A ∩ B′) (iv) P(B ∩ A′)

Solution

P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.
(i) P ( A ∪ B) = P(A) + p(B) - P( A ∩ B)
= 0.54 + 0.69 - 0.35
= 0. 88
(ii) 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) 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:

S. No.	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?

Solution

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 = 0.6 and P(F) = 2/5 = 0.4
Since there is only one male who is over 35 years of age,
So P ( E ∩ F) = 1/5 = 0.2
the probability that the spokesperson will be either male or over 35 year
= P( E ∪ F)
\[P\left( {E \cup F} \right) = P\left( E \right) + P\left( F \right) - P\left( {E \cap F} \right)\] \[ = 0.6 + 0.4 - 0.2\] \[ = 0.8\] Thus, the probability that the spokesperson will either be a male or over 35 years of age is.0.8 = 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?

Solution

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.
To form the number greater than 5000, then for thousand place we have 2 options as 5 or 7.
For the remaining places we can select any digits.
It can be done by 5 × 5 × 5 = 125 different ways.
Total possible numbers formed = 2 × 125 = 250
But number 5000 is not greater than 5000. So we will substact i from 250.
So total possible outcomes n = 250 - 1 = 249
The number must be divisible by 5, so its unit place digit must be 5 or 0. So have 2 options.
The number formed greater than 5000 and divisible by 5 = 2 × 5 × 5 × 2 = 100
But number 5000 is not greater than 5000. So we will substact i from 100.
So faverable outcomes m = 100 - 1 = 99
So probability that number is divisible by 5
\[ = \frac{m}{n}\] \[ = \frac{{99}}{{249}}\] \[ = \frac{{33}}{{83}}\] (ii) When repetition of digits is not allowed
The thousands place can be filled with either of the two digits 5 or 7.
The remaining 3 places can be filled with any of the remaining 4 digits.
∴Total number of 4-digit numbers greater than 5000 = 2 × 4 × 3 × 2
n = 48
When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining 3 digits.
∴Here, number of 4-digit numbers starting with 5 and divisible by 5 = 3 × 2 = 6
When the digit at the thousands place is 7, the units place can be filled in two ways (0 or 5) and the tens and hundreds places can be filled with any two of the remaining 3 digits.
∴ Here, number of 4-digit numbers starting with 7 and divisible by 5
= 1 × 2 × 3 × 2 = 12
∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18
Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed is
\[ = \frac{m}{n}\] \[ = \frac{{18}}{{48}}\] \[ = \frac{3}{8}\]

Question (10)

The number lock of a suitcase has 4 wheels, each labelled 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

The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats.
As repetation is not allowed, it means we have form a four digit number.
It can be done by 10P₄ different ways.
So the possible out comes n = 10P₄
= 10 × 9 × 8 × 7
= 5040.
The lock will be open when the number on wheel match with the number set by owner.
So possible outcomes to open suitcase m = 1
So probability of getting correct sequence = m/n
= 1 / 5040.