Answer :

Total number of marbles $=10+20+30=60$

Number of ways of drawing 5 marbles from 60 marbles $={ }^{60} C_5$

(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 ${ }^{20} C_5$ ways. $\therefore$ Probability that all marbles will be blue $=$

$ =\dfrac{{ }^{20} C_5}{{ }^{60} C_5} $

(ii) Number of ways in which the drawn marble is not green $={ }^{(20+10)} C_5={ }^{30} C_5$

$\therefore$ Probability that no marble is green $=\dfrac{{ }^{30} C_5}{{ }^{60} C_5}$

$\therefore$ Probability that at least one marble is green $=$

$ =1-\dfrac{{ }^{30} C_5}{{ }^{60} C_5} $

Answer :

Number of ways of drawing 4 cards from 52 cards $={ }^{52} C_4$

In a deck of 52 cards, there are 13 diamonds and 13 spades.

$\therefore$ Number of ways of drawing 3 diamonds and one spade $={ }^{13} C_3 \times{ }^{13} C_1$

Thus, the probability of obtaining 3 diamonds and one spade $=$

$ \dfrac{{ }^{13} C_3 \times{ }^{13} C_1}{{ }^{52} C_4} $

Answer :

Total number of faces $=6$

(i) Number faces with number ’ 2 ’ $=3$

$\therefore P(2)=\dfrac{3}{6}=\dfrac{1}{2}$

(ii) $P(1$ or 3$)=P($ not 2$)=1-\dfrac{1}{2}=\dfrac{1}{2}$

(iii) Number of faces with number ’ 3 ’ $=1$

$\therefore P(3)=\dfrac{1}{6}$

Thus, $P(not 3)=1-P(3)=1-\dfrac{1}{6}=\dfrac{5}{6}$

Answer :

Total number of tickets sold $=10,000$

Number prizes awarded $=10$

(i) If we buy one ticket, then

$P($ getting a prize $)=\dfrac{10}{10000}=\dfrac{1}{1000}$

$\therefore P($ not getting a prize $)=1-\dfrac{1}{1000}=\dfrac{999}{1000}$

(ii) If we buy two tickets, then

Number of tickets not awarded $=10,000-10=9990$

$P($ not getting a prize $)=$

$ \dfrac{{ }^{9990} C_2}{{ }^{10000} C_2} $

(iii) If we buy 10 tickets, then

$P($ not getting a prize $)=$

$ \dfrac{{ }^{9990} C _{10}}{{ }^{100000} C _{10}} $

Answer :

My friend and I are among the 100 students.

Total number of ways of selecting 2 students out of 100 students $={ }^{100} C_2$

(a) The two of us will enter the same section if both of us are among 40 students or among 60 students.

$\therefore$ Number of ways in which both of us enter the same section $={ }^{40} C_2+{ }^{60} C_2$

$\therefore$ Probability that both of us enter the same section

$ = \large\dfrac{{ }^{40}C_2 + { }^{60}C_2}{{ }^{100}C_2}= \large\dfrac{\large\dfrac{\lfloor{40}}{\lfloor{2}\lfloor{38}} + \large\dfrac{\lfloor{60}}{\lfloor{2}\lfloor{58}}}{\large\dfrac{\lfloor{100}}{{\lfloor{2}\lfloor{98}}}} = \dfrac{(39 \times 40 )+ (59 \times 60)}{99 \times 100} = \dfrac{17}{33} $

(b) $P$ (we enter different sections)

$=1-P$ (we enter the same section)

$ =1-\dfrac{17}{33}=\dfrac{16}{33} $

Answer :

Let $L_1, L_2, L_3$ be three letters and $E_1, E_2$, and $E_3$ be their corresponding envelops respectively.

There are 6 ways of inserting 3 letters in 3 envelops. These are as follows:

$ \begin{aligned} & L_1 E_1, L_2 E_3, L_3 E_2 \\ & L_2 E_2, L_1 E_3, L_3 E_1 \\ & L_3 E_3, L_1 E_2, L_2 E_1 \\ & L_1 E_1, L_2 E_2, L_3 E_3 \\ & L_1 E_2, L_2 E_3, L_3 E_1 \\ & L_1 E_3, L_2 E_1, L_3 E_2 \end{aligned} $

There are 4 ways in which at least one letter is inserted in a proper envelope.

Thus, the required probability is $\dfrac{4}{6}=\dfrac{2}{3}$.

Answer :

It is given that $P(A)=0.54, P(B)=0.69, P(A \cap B)=0.35$

(i) We know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\therefore P(A \cup B)=0.54+0.69-0.35=0.88$

(ii) $A^ \prime \cap B^\prime =(A \cup B)^\prime $ [by De Morgan’s law]

$\therefore P(A^\prime \cap B^\prime)=P(A \cup B)^\prime=1-P(A \cup B)=1-0.88=0.12$

(iii) $P(A \cap B )^\prime=P(A)-P(A \cap B)$

$=0.54-0.35$

$=0.19$

(iv) We know that $n(B \cap A^{\prime})=n(B)-n(A \cap B)$

$\Rightarrow \dfrac{n(B \cap A^{\prime})}{n(S)}=\dfrac{n(B)}{n(S)}-\dfrac{n(A \cap B)}{n(S)}$

$\therefore P(B \cap A^{\prime})=P(B)-P(A \cap B)$

$\therefore P(B \cap A^{\prime})=0.69-0.35=0.34$

Answer :

Let $E$ be the event in which the spokesperson will be a male and $F$ be the event in which the spokespen will be over 35 years of age.

Accordingly, $P(E)=\dfrac{3}{5}$ and $P(F)=\dfrac{2}{5}$

Since there is only one male who is over 35 years of age,

$P(E \cap F)=\dfrac{1}{5}$

We know that $P(E \cup F)=P(E)+P(F)-P(E \cap F)$

$\therefore P(E \cup F)=\dfrac{3}{5}+\dfrac{2}{5}-\dfrac{1}{5}=\dfrac{4}{5}$

Thus, the probability that the spokesperson will either be a male or over 35 years of age is $\dfrac{4}{5}$

Answer :

(i) When the digits are repeated

Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5 .

The remaining 3 places can be filled by any of the digits $0,1,3,5$, or 7 as repetition of digits is allowed.

$\therefore$ Total number of 4-digit numbers greater than $5000=2 \times 5 \times 5 \times 5$ - 1 $=250-1=249$

[In this case, 5000 can not be counted; so 1 is subtracted]

A number is divisible by 5 if the digit at its units place is either 0 or 5 .

$\therefore$ Total number of 4-digit numbers greater than 5000 that are divisible by $5=2 \times 5 \times 5 \times 2-1=100-1=99$

Thus, the probability of forming a number divisible by 5 when the digits are repeated is

$ =\dfrac{99}{249}=\dfrac{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.

$\therefore$ Total number of 4-digit numbers greater than $5000=2 \times 4 \times 3 \times 2$

$=48$

When the digit at the thousands place is 5 , the units place can be filled only with 0 and the tens and hundreds can be filled with any two of the remaining 3 digits.

$\therefore$ Here, number of 4-digit numbers starting with 5 and divisible by 5

$=3 \times 2=6$

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.

$\therefore$ Here, number of 4-digit numbers starting with 7 and divisible by 5

$=1 \times 2 \times 3 \times 2=12$

$\therefore$ 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 $\dfrac{18}{48}=\dfrac{3}{8}$.

Answer :

The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9.

Number of ways of selecting 4 different digits out of the 10 digits $={ }^{10} C_4$

Now, each combination of 4 different digits can be arranged in 4 ways.

$\therefore$ Number of four digits with no repetitions $={ }^{10} C_4 \times\lfloor 4=\dfrac{\lfloor 10}{\lfloor 46} \times 44=7 \times 8 \times 9 \times 10=5040.$

There is only one number that can open the suitcase.

Thus, the required probability is $\dfrac{1}{5040}$