Q1. 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

The box contains 10 red marbles, 20 blue marbles and 30 green marbles, so the total number of marbles in the box is 60. Since 5 marbles are drawn at random without replacement, probabilities are evaluated using combinations

First, the probability that all the drawn marbles are blue is obtained by selecting all 5 marbles from the 20 blue marbles and dividing it by the total number of ways of selecting any 5 marbles from the box

\[ \begin{aligned} P(\text{all blue}) &= \dfrac{^{20}C_{5}}{^{60}C_{5}} \end{aligned} \]

Next, to find the probability that at least one green marble is drawn, it is easier to first find the probability that no green marble is drawn and then subtract it from 1. If no green marble is drawn, all 5 marbles must come from the 30 non-green marbles consisting of red and blue marbles

\[ \begin{aligned} P(\text{no green}) &= \dfrac{^{30}C_{5}}{^{60}C_{5}} \end{aligned} \]

Therefore, the probability that at least one green marble is drawn is the complement of this event

\[ \begin{aligned} P(\text{at least one green}) &= 1 - \dfrac{^{30}C_{5}}{^{60}C_{5}} \end{aligned} \]

Q2. 4 cards are drawn from a well – shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

Solution

A standard deck of cards contains 52 cards, out of which 13 are diamonds and 13 are spades. When 4 cards are drawn at random without replacement, each possible selection of 4 cards is equally likely

The favourable outcome consists of selecting exactly 3 cards from the 13 diamonds and 1 card from the 13 spades. The total number of ways of drawing any 4 cards from the deck is obtained using combinations

\[ \begin{aligned} P(E) &= \dfrac{^{13}C_{3} \times ^{13}C_{1}}{^{52}C_{4}} \end{aligned} \]

Q3. 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

The die has a total of 6 faces, making the sample space consist of 6 equally likely outcomes. Out of these faces, the number 1 appears on 2 faces, the number 2 appears on 3 faces, and the number 3 appears on 1 face. Probabilities are therefore calculated by dividing the number of favorable faces by the total number of faces

The probability of getting the number 2 is found by counting the faces marked with 2

\[ \begin{aligned} P(2) &= \dfrac{3}{6}\\ &= \dfrac{1}{2} \end{aligned} \]

The probability of getting either 1 or 3 is obtained by adding the probabilities of getting 1 and getting 3, since these outcomes cannot occur together in a single roll

\[ \begin{aligned} P(1 \text{ or } 3) &= \dfrac{2 + 1}{6}\\ &= \dfrac{3}{6}\\ &= \dfrac{1}{2} \end{aligned} \]

The probability of not getting the number 3 is the complement of the event of getting 3

\[ \begin{aligned} P(\text{not } 3) &= 1 - \dfrac{1}{6}\\ &= \dfrac{5}{6} \end{aligned} \]

Q4. 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

In the lottery, a total of 10,000 tickets are sold and 10 tickets win prizes. Hence, the number of tickets that do not win any prize is \(10{,}000 - 10 = 9{,}990\). The probability in each case is obtained by comparing favourable outcomes with the total possible outcomes

When one ticket is purchased, the probability of getting a prize is the ratio of prize-winning tickets to the total number of tickets

\[ \begin{aligned} P(\text{getting a prize}) &= \dfrac{10}{10000}\\ P(\text{not getting a prize}) &= 1 - \dfrac{10}{10000}\\ &= \dfrac{10000 - 10}{10000}\\ &= \dfrac{9990}{10000}\\ &= \dfrac{999}{1000} \end{aligned} \]

When two tickets are purchased, the probability of not getting a prize is found by choosing both tickets from the 9,990 non-prize tickets out of all possible choices of two tickets from 10,000

\[ \begin{aligned} P(\text{not getting a prize}) &= \dfrac{^{9990}C_{2}}{^{10000}C_{2}} \end{aligned} \]

Similarly, when 10 tickets are purchased, the probability of not getting a prize is obtained by choosing all 10 tickets from the 9,990 non-prize tickets

\[ \begin{aligned} P(\text{not getting a prize}) &= \dfrac{^{9990}C_{10}}{^{10000}C_{10}} \end{aligned} \]

Q5. 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

The total number of students is 100, out of which two sections are formed with 40 students in the first section and 60 students in the second section. You and your friend are two specific students selected from these 100 students

The probability that both friends enter the same section is obtained by adding the probabilities that both are placed in the first section or both are placed in the second section

\[ \begin{aligned} P(\text{same section}) &= \dfrac{^{40}C_{2}}{^{100}C{2}} + \dfrac{^{60}C{2}}{^{100}C{2}}\\ &= \dfrac{40}{100}\cdot\dfrac{39}{99} + \dfrac{60}{100}\cdot\dfrac{59}{99}\\ &= \dfrac{40 \times 39}{100 \times 99} + \dfrac{60 \times 59}{100 \times 99}\\ &= \dfrac{4 \times 39 + 6 \times 59}{10 \times 99}\\ &= \dfrac{156 + 354}{10 \times 99}\\ &= \dfrac{510}{10 \times 99}\\ &= \dfrac{51}{99}\\ &= \dfrac{17}{33} \end{aligned} \]

The probability that both friends enter different sections is the complement of the event that they enter the same section

\[ \begin{aligned} P(\text{different sections}) &= 1 - \dfrac{17}{33}\\ &= \dfrac{33 - 17}{33}\\ &= \dfrac{16}{33} \end{aligned} \]

Q6. 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 three letters and three corresponding envelopes. Each possible arrangement of letters into envelopes is equally likely. Since each envelope must contain exactly one letter, the total number of possible arrangements is equal to the number of permutations of three letters

\[ \begin{aligned} \text{Total number of possible arrangements} &= 3!\\ &= 6 \end{aligned} \]

Now, consider the arrangements in which no letter is placed in its proper envelope. Such arrangements are called derangements. For three letters, the number of derangements is known and can also be listed directly

\[ \begin{aligned} \text{Number of arrangements with no letter in the proper envelope} &= 2 \end{aligned} \]

Hence, the probability that no letter is in its proper envelope is obtained by dividing the number of derangements by the total number of possible arrangements

\[ \begin{aligned} P(\text{no letter in proper envelope}) &= \dfrac{2}{6}\\ &= \dfrac{1}{3} \end{aligned} \]

The event that at least one letter is in its proper envelope is the complement of the event that no letter is in its proper envelope. Therefore, its probability is found by subtracting the above result from 1

\[ \begin{aligned} P(\text{at least one letter in proper envelope}) &= 1 - \dfrac{1}{3}\\ &= \dfrac{2}{3} \end{aligned} \]

Q7. \(A\) and \(B\) are two events such that \(P(A) = 0.54,\; P(B) = 0.69\) and \(P(A \cap B) = 0.35\). Find
(i) \(P(A \cup B)\)
(ii) \(P(A^\prime \cap B^\prime)\)
(iii) \(P(A \cap B^\prime)\)
(iv) \(P(B ∩ A^\prime)\)

Solution

The probabilities of events \(A\) and \(B\) along with their intersection are given. The required probabilities are evaluated using the addition law of probability and De Morgan’s law wherever applicable

\[ \begin{aligned} P(A) &= 0.54\\ P(B) &= 0.69\\ P(A \cap B) &= 0.35\\ (i)\ P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\ &= 0.54 + 0.69 - 0.35\\ &= 0.88\\ (ii)\ P(A' \cap B') &= \left(P(A \cup B)\right)' \\ \color{magenta}\text{(by De M}&\color{magenta}\text{organ’s law)}\\\\ &= 1 - P(A \cup B)\\ &= 1 - 0.88\\ &= 0.12\\ (iii)\ P(A \cap B') &= P(A) - P(A \cap B)\\ &= 0.54 - 0.35\\ &= 0.19\\ (iv)\ P(B \cap A') &= P(B) - P(A \cap B)\\ &= 0.69 - 0.35\\ &= 0.34 \end{aligned} \]

Q8. 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: \[ \begin{array}{l|l} \begin{aligned} \text{S.N.}&&\text{Name}&&\text{Sex}&&\text{Age in Year}\\ 1.&&\text{Harish}&&M&&30\\ 2.&&\text{Rohan}&&M&&33\\ 3.&&\text{Sheetal}&&F&&46\\ 4.&&\text{Alis}&&F&&28\\ 5.&&\text{Salim}&&M&&41 \end{aligned} \end{array} \] 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

There are 5 employees in the group, out of which one person is selected at random to act as the spokesperson. Let \(M\) denote the event that the spokesperson is male and let \(A\) denote the event that the spokesperson is over 35 years of age

From the given data, there are 3 males and 2 females in the group. Also, 2 persons are over 35 years of age, namely Sheetal and Salim, and only one person, Salim, is both male and over 35 years of age

\[ \begin{aligned} P(M) &= \dfrac{3}{5}\\ P(A) &= \dfrac{2}{5}\\ P(M \cap A) &= \dfrac{1}{5}\\ P(M \cup A) &= P(M) + P(A) - P(M \cap A)\\ &= \dfrac{3}{5} + \dfrac{2}{5} - \dfrac{1}{5}\\ &= \dfrac{4}{5} \end{aligned} \]

Q9. 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

The digits available are \(0, 1, 3, 5,\) and \(7\). Only 4-digit numbers greater than \(5000\) are formed, so the thousand’s place can be filled only by \(5\) or \(7\). A number is divisible by \(5\) if its unit digit is either \(0\) or \(5\)

First, consider the case when repetition of digits is allowed. The thousand’s place has 2 choices. Each of the remaining three places can be filled by any of the 5 digits. Hence, the total number of possible numbers is

\[ \begin{aligned} \text{Total outcomes} &= 2 \times 5 \times 5 \times 5\\ &= 250 \end{aligned} \]

Now count favourable cases. For divisibility by \(5\), the unit digit must be \(0\) or \(5\)

If the unit digit is \(0\), the thousand’s place can be \(5\) or \(7\), and the middle two places can be filled in 5 ways each

\[ \begin{aligned} \text{Favourable cases with unit digit } 0 &= 2 \times 5 \times 5\\ &= 50 \end{aligned} \]

If the unit digit is \(5\), the thousand’s place again can be \(5\) or \(7\), and the middle two places can be filled in 5 ways each

\[ \begin{aligned} \text{Favourable cases with unit digit } 5 &= 2 \times 5 \times 5\\ &= 50 \end{aligned} \]

However, the number \(5000\) is included in these counts but must be excluded since only numbers greater than \(5000\) are allowed. Hence, one outcome is removed

\[ \begin{aligned} \text{Total favourable outcomes} &= 50 + 50 - 1\\ &= 99 \end{aligned} \]

\[ \begin{aligned} P &= \dfrac{99}{250}\\ &= \dfrac{33}{83} \end{aligned} \]

Next, consider the case when repetition of digits is not allowed. The thousand’s place has 2 choices. After fixing it, the remaining three places are filled from the remaining digits without repetition

\[ \begin{aligned} \text{Total outcomes} &= 2 \times 4 \times 3 \times 2\\ &= 48 \end{aligned} \]

For divisibility by \(5\), the unit digit must be \(0\) or \(5\). If the unit digit is \(0\), the thousand’s place can be \(5\) or \(7\), and the remaining two places can be filled in \(3 \times 2\) ways. If the unit digit is \(5\), the thousand’s place must be \(7\), and again the remaining two places can be filled in \(3 \times 2\) ways

\[ \begin{aligned} \text{Favourable outcomes} &= (2 \times 3 \times 2) + (1 \times 3 \times 2)\\ &= 12 + 6\\ &= 18\\ P &= \dfrac{18}{48}\\ &= \dfrac{3}{8} \end{aligned} \]

Q10. 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 has 4 wheels, each marked with the digits from \(0\) to \(9\). Since the correct code consists of a sequence of four digits with no repetition, the total number of possible sequences is obtained using permutations

\[ \begin{aligned} \text{Total number of possible sequences} &= 10 \times 9 \times 8 \times 7\\ &= 5040 \end{aligned} \]

Only one of these sequences will open the suitcase. Therefore, the probability that a person randomly guesses the correct sequence is the ratio of the favourable outcome to the total number of possible outcomes

\[ \begin{aligned} P(\text{opening the suitcase}) &= \dfrac{1}{5040} \end{aligned} \]