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Maths (commerce) Solve the Following Question.(2 Marks)

Probability (p-2) · Maharashtra Board STD 11 Commerce / Arts (MSBSHSE - English Medium). 33 questions.

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Solve the Following Question.(2 Marks)

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33 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
The odds against a certain event are $5: 2$ and the odds in favour of another independent event are $6: 5$. Find the chance that at least one of the events will happen.
Answer
Let $A$ and $B$ be two independent events.
Odds against $A$ are $5: 2$
$\therefore$ the probability of occurrence of event $A$ is given by
$
\mathrm{P}(\mathrm{A})=\frac{2}{5+2}=\frac{2}{7}
$
Odds in favour of $B$ are $6: 5$
$\therefore$ the probability of occurrence of event $B$ is given by
$
\begin{aligned}
& P(B)=\frac{6}{6+5}=\frac{6}{11} \\
& \therefore P(\text { at least one event will happen })=P(A \cup B) \\
& =P(A)+P(B)-P(A \cap B) \\
& =P(A)+P(B)-P(A) P(B) \ldots . . .[\because A \text { and } B \text { are independent events }] \\
& =\frac{2}{7}+\frac{6}{11}-\frac{2}{7} \times \frac{6}{11} \\
& =\frac{2}{7}+\frac{6}{11}-\frac{12}{77}=\frac{22+42-12}{77}=\frac{52}{77}
\end{aligned}
$
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Question 22 Marks
$A, B$ and $C$ are mutually exclusive and exhaustive events associated with the random experiment. Find $P(A)$, given that
$P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$
Answer
$P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$
Since $A, B, C$ are mutually exclusive and exhaustive events,
$
\begin{array}{ll}
\therefore & \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})=1 \\
\therefore & \mathrm{P}(\mathrm{A})+\frac{3}{2} \mathrm{P}(\mathrm{A})+\frac{1}{2} \mathrm{P}(\mathrm{B})=1 \\
\therefore & \mathrm{P}(\mathrm{A})+\frac{3}{2} \mathrm{P}(\mathrm{A})+\frac{1}{2} \times \frac{3}{2} \mathrm{P}(\mathrm{A})=1 \\
\therefore & \mathrm{P}(\mathrm{A})+\frac{3}{2} \mathrm{P}(\mathrm{A})+\frac{3}{4} \mathrm{P}(\mathrm{A})=1 \\
\therefore & \mathrm{P}(\mathrm{A}) \times\left(1+\frac{3}{2}+\frac{3}{4}\right)=1 \\
\therefore & \mathrm{P}(\mathrm{A}) \times\left(\frac{13}{4}\right)=1 \\
\therefore & \mathrm{P}(\mathrm{A})=\frac{4}{13}
\end{array}
$
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Question 32 Marks
A box contains 25 tickets numbered 1 to 25 . Two tickets are drawn at random. What is the probability that the product of the numbers is even?
Answer
Two tickets can be drawn out of 25 tickets in ${ }^{25} \mathrm{C}_2=\frac{25 \times 24}{1 \times 2}=300$ ways.
$
\therefore \mathrm{n}(\mathrm{S})=300
$
Let $\mathrm{A}$ be the event that product of two numbers is even.
This is possible if both numbers are even, or one number is even and other is odd. As there are 13 odd numbers and 12 even numbers from 1 to 25 .
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})={ }^{12} \mathrm{C}_2+{ }^{12} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \\
& =\frac{12 \times 11}{1 \times 2}+12 \times 13 \\
& =66+156 \\
& =222 \\
& \therefore \text { Required probability }=\mathrm{P}(\mathrm{A}) \\
& =\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})} \\
& =\frac{222}{300} \\
& =\frac{37}{50}
\end{aligned}
$
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Question 42 Marks
A number of two digits is formed using the digits $1,2,3, \ldots . . ., 9$. What is the probability that the number so chosen is even and less than 60 ?
Answer
The number of two digits can be formed from the given 9 digits in $9 \times 9=81$ different ways.
$
\therefore \mathrm{n}(\mathrm{S})=81
$
Let $\mathrm{A}$ be the event that the number is even and less than 60 .
Since the number is even, the unit place of two digits can be filled in ${ }^4 \mathrm{P}_1=4$ different ways by any one of digits $2,4,6,8$.
Also the number is less than 60 , so tenth place can be filled in ${ }^5 P_1=5$ different ways by any one of the digits $1,2,3,4,5$.
$
\therefore \mathrm{n}(\mathrm{A})=4 \times 5=20
$
$\therefore$ Required probability $=P(A)=\frac{n(A)}{n(S)}=\frac{20}{81}$
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Question 52 Marks
Two-thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting first class is 0.25 and that of a boy getting is 0.28 . Find the probability that a student chosen at random will get first class.
Answer
Let $\mathrm{A}$ be the event that student chosen is a boy
$\mathrm{B}$ be the event that student chosen is a girl
$\mathrm{C}$ be the event that student gets first class
$
\therefore \mathrm{P}(\mathrm{A})=\frac{2}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{3}
$
Probability of student getting first class, given that student is boy
Probability of student getting first class given that student is a girl, is
$
P(C / A)=0.28=\frac{28}{100}
$
and $P(C / B)=0.25=\frac{25}{100}$
$\therefore$ Required probability $=\mathrm{P}((\mathrm{A} \cap \mathrm{C}) \cup(B \cap C))$
Since $A \cap C$ and $B \cap C$ are mutually exclusive events
$\therefore$ Required probability $=\mathrm{P}(\mathrm{A} \cap \mathrm{C})+\mathrm{P}(\mathrm{B} \cap \mathrm{C})$
$
\begin{aligned}
& =P(A) \cdot P(C / A)+P(B) \cdot P(C / B) \\
& =\left(\frac{2}{3} \times \frac{28}{100}\right)+\left(\frac{1}{3} \times \frac{25}{100}\right) \\
& =\frac{56+25}{300}=\frac{81}{300}=\frac{27}{100}=0.27
\end{aligned}
$
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Question 62 Marks
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that
(i) first is white and second is black?
(ii) one is white and the other is black?
Answer
Total number of balls $=10+15=25$
Let $\mathrm{S}$ be an event that two balls are drawn at random without replacement in succession
$
\therefore \mathrm{n}(\mathrm{S})={ }^{25} \mathrm{C}_1 \times{ }^{24} \mathrm{C}_1=25 \times 24
$
(i) Let $\mathrm{A}$ be the event that the first ball is white and the second is black.
First white ball can be drawn from 10 white balls in ${ }^{10} \mathrm{C}_1$ ways and second black ball can be drawn from 15 black balls in ${ }^{15} \mathrm{C}_1$ ways.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})={ }^{10} \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{10} \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1}{25 \times 24}=\frac{10 \times 15}{25 \times 24}=\frac{1}{4} \\
& \therefore \quad \mathrm{n}(\mathrm{B})={ }^{10} \mathrm{C}_1 \cdot{ }^{15} \mathrm{C}_1+{ }^{15} \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_1 \\
& \therefore \quad \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{10 \times 15}{25 \times 24}+\frac{15 \times 10}{25 \times 24} \\
& =\frac{150}{25 \times 24}+\frac{150}{25 \times 24} \\
& =\frac{300}{25 \times 24}=\frac{1}{2}
\end{aligned}
$
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Question 72 Marks
One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third gun respectively. Assuming that $A, B$, and $C$ are independent events and that $P(A)=0.5, P(B)=0.6$, and $P(C)=0.8$, then find the probability that at least one hit is registered.
Answer
A be the event that first gun hits the target
$B$ be the event that second gun hits the target
$\mathrm{C}$ be the event that third gun hits the target
$
\begin{aligned}
& P(A)=0.5, P(B)=0.6, P(C)=0.8 \\
& \therefore P\left(A^{\prime}\right)=1-P(A)=1-0.5=0.5 \\
& \therefore P\left(B^{\prime}\right)=1-P(B)=1-0.6=0.4 \\
& \therefore P\left(C^{\prime}\right)=1-P(C)=1-0.8=0.2
\end{aligned}
$
Now $A, B, C$ are independent events
$\therefore A^{\prime}, B^{\prime}, C$ are also independent events.
$\therefore \mathrm{P}$ (at least one hit is registered)
$=1-\mathrm{P}$ (no hit is registered)
$=1-P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)$
$=1-P\left(A^{\prime}\right) P\left(B^{\prime}\right) P\left(C^{\prime}\right)$
$=1-(0.5)(0.4)(0.2)$
$=1-0.04$
$=0.96$
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Question 82 Marks
The probability that a 50 -year old man will be alive till age 60 is 0.83 and the probability that a 45 -year old woman will be alive till age 55 is 0.97 . What is the probability that a man whose age is 50 and his wife whose age is 45 will both be alive for the next 10 years?
Answer
Let $\mathrm{A}$ be the event that man will be alive at 60 .
$
\therefore \mathrm{P}(\mathrm{A})=0.83
$
Let $B$ be the event that a woman will be alive at 55 .
$
\therefore \mathrm{P}(\mathrm{B})=0.97
$
$A \cap B=$ Event that both will be alive.
Also, $A$ and $B$ are independent events
$
\begin{aligned}
& \therefore P(\text { both man and his wife will be alive })=P(A \cap B) \\
& =P(A) \cdot P(B) \\
& =0.83 \times 0.97 \\
& =0.8051
\end{aligned}
$
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Question 92 Marks
A bag contains 4 blue and 5 green balls. Another bag contains 3 blue and 7 green balls. If one ball is drawn from each bag, what is the Probability that two balls are of the same colour?
Answer
Let $\mathrm{A}$ be the event that a blue ball is drawn from each bag.
Probability of drawing one blue ball out of 4 blue balls where there are a total of 9 balls in the first bag and that of drawing one blue ball out of 3 blue balls where there are a total of 10 balls in the second bag is
$
P(A)=\frac{4}{9} \times \frac{3}{10}
$
Let $B$ be the event that a green ball is drawn from each bag.
Probability of drawing one green ball out of 5 green balls where there are a total of 9 balls in the first bag and that of drawing one green ball out of 7 green balls where there are a total of 10 balls in the second bag is
$
P(B)=\frac{5}{9} \times \frac{7}{10}
$
Since both, the events are mutually exclusive and exhaustive events
$\therefore \mathrm{P}($ that both the balls are of the same colour $)=\mathrm{P}($ both are of blue colour $)$ or $\mathrm{P}($ both are of green colour)
$
\begin{aligned}
& =P(A)+P(B) \\
& =\frac{4}{9} \times \frac{3}{10}+\frac{5}{9} \times \frac{7}{10} \\
& =\frac{12}{90}+\frac{35}{90} \\
& =\frac{47}{90}
\end{aligned}
$
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Question 102 Marks
Two balls are drawn from an urn containing 5 green, 3 blue, 7 yellow balls one by one without replacement. What is the probability that at least one ball is blue?
Answer
Total number of balls in the urn $=5+3+7=15$
Out of these 12 are non-blue balls.
Two balls can be drawn from 15 balls without replacement in ${ }^{15} \mathrm{C}_2$
$
\begin{aligned}
& =\frac{\mathbf{1 5 \times 1 4}}{\mathbf{1} \times 2} \\
& =105 \text { ways. } \\
& \therefore \mathrm{n}(\mathrm{S})=105
\end{aligned}
$
Let $\mathrm{A}$ be the event that at least one ball is blue, i.e., 1 blue and other non-blue or both are blue.
$
\begin{aligned}
& \therefore n(A)={ }^3 C_1 \times{ }^{12} C_1+{ }^3 C_2 \\
& =3 \times 12+3 \\
& =36+3 \\
& =39 \\
& \therefore P(A)=\frac{n(A)}{n(S)}=\frac{39}{105}=\frac{13}{35}
\end{aligned}
$
Alternate solution:
Total number of balls in the urn $=15$
Required probability $=1-\mathrm{P}$ (neither of two balls is blue)
Balls are drawn one by one without replacement.
Probability of first non-blue ball drawn $=\frac{12}{15}$
Probability of second non-blue ball drawn $=\frac{11}{14}$
Probability of neither of two ball is blue $=\frac{12}{15} \times \frac{11}{14}=\frac{22}{35}$
$\therefore$ Required probability $=1-\frac{22}{35}=\frac{13}{35}$
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Question 112 Marks
An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, what is the probability that at least one ball is black?
Answer
Total number of balls in the urn $=4+5+6=15$
Two balls can be drawn without replacement in ${ }^{15} \mathrm{C}_2=\frac{15 \times 14}{1 \times 2}=105$ ways
$
\therefore \mathrm{n}(\mathrm{S})=105
$
Let $\mathrm{A}$ be the event that at least one ball is black
i.e., 1 black and 1 non-black or 2 black and 0 non-black.
1 black ball can be drawn out of 4 black balls in ${ }^4 C_1=4$ ways
and 1 non-black ball can be drawn out of remaining 11 non-black balls in ${ }^{11} C_1=11$ ways
$\therefore 1$ black and 1 non black ball can be drawn in $4 \times 11=44$ ways
Also, 2 black balls can be drawn from 4 black balls in ${ }^4 C_2=\frac{4 \times 3}{1 \times 2}=6$ ways
$\therefore \mathrm{n}(\mathrm{A})=44+6=50$
$\therefore$ Required probability $=\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{50}{105}=\frac{10}{21}$
Alternate Solution:
Total number of balls $=15$
Required probability $=1-\mathrm{P}$ (neither of two balls is black)
Balls are drawn without replacement
Probability of first non-black ball drawn $=\frac{11}{15}$
Probability of second non-black ball drawn $=\frac{10}{14}$
Probability of neither of two balls is black $=\frac{11}{15} \times \frac{10}{14}=\frac{11}{21}$
Required probability $=1-\frac{11}{21}=\frac{10}{21}$
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Question 122 Marks
A computer software company is bidding for computer programs A and B. The probability that the company will get software $A$ is $\frac{3}{5}$, the probability that the company will get software $B$ is $\frac{1}{3}$ and the probability that company will get both $A$ and $B$ is $\frac{1}{8}$. What is the probability that the company will get at least one software?
Answer
Let $\mathrm{A}$ be the event that the company will get software $\mathrm{A}$.
$
\therefore P(A)=\frac{3}{5}
$
Let $B$ be the event that company will get software $B$.
$
\therefore P(B)=\frac{1}{3}
$
Also, $P(A \cap B)=\frac{1}{8}$
$
\begin{aligned}
& \therefore P(\text { the company will get at least one software })=P(A \cup B) \\
& =P(A)+P(B)-P(A \cap B) \\
& =\frac{3}{5}+\frac{1}{3}-\frac{1}{8} \\
& =\frac{72+40-15}{120} \\
& =\frac{97}{120}
\end{aligned}
$
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Question 132 Marks
A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random: failed in both examinations.
Answer
Out of hundred students 1 student can be selected in ${ }^{100} \mathrm{C}_1=100$ ways.
$
\therefore \mathrm{n}(\mathrm{S})=100
$
Let $\mathrm{A}$ be the event that the student passed in the first examination.
Let $B$ be the event that student passed in second examination.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})=60, \mathrm{n}(\mathrm{B})=50 \text { and } \mathrm{n}(\mathrm{A} \cap \mathrm{B})=30 \\
& \therefore \quad \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{60}{100}=\frac{6}{10} \\
& \therefore \quad \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{50}{100}=\frac{5}{10} \\
& \\
& \therefore \quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{30}{100}=\frac{3}{10} \\
&
\end{aligned}
$
$
\begin{aligned}:\mathrm{P}(\text { student failed in both examinations })=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right) \\
& =\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime} \text {.....[De Morgan's law] } \\
& =1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\
& =1-\frac{4}{5} \\
& =\frac{1}{5}
\end{aligned}
$
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Question 142 Marks
A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random: passed in exactly one examination.
Answer
Out of hundred students 1 student can be selected in ${ }^{100} \mathrm{C}_1=100$ ways.
$
\therefore \mathrm{n}(\mathrm{S})=100
$
Let $\mathrm{A}$ be the event that the student passed in the first examination.
Let $B$ be the event that student passed in second examination.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})=60, \mathrm{n}(\mathrm{B})=50 \text { and } \mathrm{n}(\mathrm{A} \cap \mathrm{B})=30 \\
& \therefore \quad \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{60}{100}=\frac{6}{10} \\
& \therefore \quad \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{50}{100}=\frac{5}{10} \\
& \\
& \therefore \quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{30}{100}=\frac{3}{10} \\
&
\end{aligned}
$
$
\begin{aligned}: $\mathrm{P}$ (student passed in exactly one examination $)=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-2 \cdot \mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$
\begin{aligned}
& =\frac{6}{10}+\frac{5}{10}-2\left(\frac{3}{10}\right) \\
& =\frac{1}{2}
\end{aligned}
$
$
\begin{aligned}
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Question 152 Marks
A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random: passed at least one examination.
Answer
Out of hundred students 1 student can be selected in ${ }^{100} \mathrm{C}_1=100$ ways.
$
\therefore \mathrm{n}(\mathrm{S})=100
$
Let $\mathrm{A}$ be the event that the student passed in the first examination.
Let $B$ be the event that student passed in second examination.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})=60, \mathrm{n}(\mathrm{B})=50 \text { and } \mathrm{n}(\mathrm{A} \cap \mathrm{B})=30 \\
& \therefore \quad \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{60}{100}=\frac{6}{10} \\
& \therefore \quad \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{50}{100}=\frac{5}{10} \\
& \\
& \therefore \quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{30}{100}=\frac{3}{10} \\
&
\end{aligned}
$
$
\begin{aligned}: P(\text { student passed in at least one examination })=P(A \cup B) \\
& =P(A)+P(B)-P(A \cap B) \\
& =\frac{6}{10}+\frac{5}{10}-\frac{3}{10} \\
& =\frac{4}{5}
\end{aligned}
$
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Question 162 Marks
The letters of the word LOGARTHM are arranged at random. Find the probability that: Start with a vowel and ends with a consonant.
Answer
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in ${ }^9 \mathrm{P}_9=9$ ! ways.
$
\therefore \mathrm{n}(\mathrm{S})=9 \text { ! }
$: Let $E$ be the event that word starts with vowel and ends with consonant.
There are 3 vowels and 6 consonants in the word LOGARTHM.
$\therefore$ The first place can be filled in 3 different ways and the last place can be filled in 6 ways.
Now, remaining 7 letters can be arranged in 7 places in ${ }^7 p_7=7$ ! ways
$
\begin{aligned}
& \therefore n(E)=3 \times 6 \times 7 ! \\
& \therefore P(E)=\frac{n(E)}{n(S)}=\frac{3 \times 6 \times 7 !}{9 !}
\end{aligned}
$
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Question 172 Marks
The letters of the word LOGARTHM are arranged at random. Find the probability that: begins with $O$ and ends with $T$
Answer
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in ${ }^9 \mathrm{P}_9=9$ ! ways.
$
\therefore \mathrm{n}(\mathrm{S})=9 \text { ! }
$: Let $D$ be the event that word begins with $O$ and ends with $T$.
Thus first and last letter can be arranged in one way each and the remaining 7 letters can be arranged in remaining 7 places in $7 p_7=7$ ! ways
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{D})=7 ! \times 1 \times 1=7 ! \\
& \therefore \mathrm{P}(\mathrm{D})=\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{s})}=\frac{7}{9 !}
\end{aligned}
$
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Question 182 Marks
The letters of the word LOGARTHM are arranged at random. Find the probability that: Exactly 4 letters between $G$ and $H$
Answer
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in ${ }^9 \mathrm{P}_9=9$ ! ways.
$
\therefore \mathrm{n}(\mathrm{S})=9 \text { ! }
$: Le: $C$ be the event that exactly 4 letters are arranged between $G$ and $H$.
Consider the following arrangement
$
123456789
$
$\therefore$ Out of 9 places, $G$ and $\mathrm{H}$ can occupy any one of following 4 positions in 4 ways.
1st and 6 th, 2 nd and 7 th, 3 rd and 8 th, 4 th and 9 th
Now, $G$ and $\mathrm{H}$ can be arranged among themselves in ${ }^2 P_2=2 !=2$ ways.
Also, the remaining 7 letters can be arranged in remaining 7 places in ${ }^7 P_7=7$ ! ways.
$
\therefore \mathrm{n}(\mathrm{C})=4 \times 2 \times 7 !=8 \times 7 !=8 \text { ! }
$
$
\therefore \mathrm{P}(\mathrm{C})=\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{8})}=\frac{\mathrm{s} !}{9 !}=\frac{\mathrm{s!}}{9 \times 8 !}=\frac{1}{9}
$
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Question 192 Marks
The letters of the word LOGARTHM are arranged at random. Find the probability that: Vowels are never together.
Answer
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in ${ }^9 \mathrm{P}_9=9$ ! ways.
$
\therefore \mathrm{n}(\mathrm{S})=9 \text { ! }
$: Let $B$ be the event that vowels are never together.
Consider the following arrangement
_c.c.c.c.c.
6 consonants create 7 gaps.
$\therefore 3$ vowels can be arranged in 7 gaps in ${ }^7 P_3$ ways.
Also 6 consonants can be arranged among themselves in fip ${ }_6=6$ ! ways.
$\therefore \mathrm{n}(\mathrm{B})=6 ! \times{ }^7 \mathrm{P}_3$
$\therefore P(B)=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{B})}=\frac{G: \mathrm{x}^{\mathrm{T}} \mathrm{P}_2}{\mathscr{T}}$
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Question 202 Marks
The letters of the word LOGARTHM are arranged at random. Find the probability that: Vowels are always together.
Answer
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in ${ }^9 \mathrm{P}_9=9$ ! ways.
$
\therefore \mathrm{n}(\mathrm{S})=9 \text { ! }
$: Let $A$ be the event that vowels are always together.
The word LOGARITHM consists of 3 vowels (O, A, D and 6 consonants (L, G, R, T, H, M). 3 vowels can be arranged among themselves in $={ }^3 p_3=3$ ! ways.
Considering 3 vowels as one group, 6 consonants and this group (i.e., altogether 7 ) can be arranged in ${ }^7 p_7=7$ ! ways.
$
\begin{aligned}
& \therefore n(A)=3 ! \times 7 ! \\
& \therefore P(A)=\frac{n(A)}{n(B)}=\frac{31 \times 7 !}{9 !}
\end{aligned}
$
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Question 212 Marks
A bag contains 15 balls of three different colours: Green, Black, and Yellow. A ball is drawn at random from the bag. The probability of a green ball is $1 / 3$. The probability of yellow is $1 / 5$.
How many balls are green, black, and yellow?
Answer
Total number of balls $=15$ and
$
P(G)=\frac{1}{3}, P(B)=\frac{7}{15}, P(Y)=\frac{1}{5}
$
$\therefore$ number of green balls $=\frac{1}{3} \times 15=5$
number of black balls $=\frac{7}{15} \times 15=7$
and number of yellow balls $=\frac{1}{5} \times 15=3$.
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Question 222 Marks
A bag contains 15 balls of three different colours: Green, Black, and Yellow. A ball is drawn at random from the bag. The probability of a green ball is $1 / 3$. The probability of yellow is $1 / 5$.
What is the probability of blackball?
Answer
The bag contains 15 balls of three different colours i.e., green (G), black (B) and yellow ( $Y$ )
$
\therefore P(G)=\frac{1}{3} \text { and } P(Y)=\frac{1}{5}
$
If a ball is drawn from the bag, then it can be any one of the green, black and yellow.
$
\begin{aligned}
& \therefore P(G)+P(B)+P(Y)=1 \\
& \therefore \frac{1}{3}+P(B)+\frac{1}{5}=1 \\
& \therefore P(B)+\frac{8}{15}=1 \\
& \therefore P(B)=1-\frac{8}{15}=\frac{7}{15}
\end{aligned}
$
$\therefore$ Probability of black ball is $\frac{7}{15}$
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Question 232 Marks
Four cards are drawn from a pack of 52 cards. Find the probability that: all cards are club and one of them is a jack.
Answer
4 cards can be drawn out of 52 cards in ${ }^{52} \mathrm{C}_4$ ways.
$
\therefore \mathrm{n}(\mathrm{S})={ }^{52} \mathrm{C}_4
$
Let $\mathrm{D}$ be the event that all the 4 cards drawn are clubs and one of them is a jack. In a pack of 52 cards, there are 13 club cards having 1 jack card.
$\therefore 1$ jack can be drawn in ${ }^1 C_1$ way and the other 3 cards can be drawn from remaining 12 club cards in ${ }^{12} \mathrm{C}_3$ ways.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{D})={ }^{12} \mathrm{C}_3 \times{ }^1 \mathrm{C}_1 \\
& \therefore \mathrm{P}(\mathrm{D})=\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_3 \times{ }^1 \mathrm{C}_1}{{ }^{32} \mathrm{C}_4}
\end{aligned}
$
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Question 242 Marks
Four cards are drawn from a pack of 52 cards. Find the probability that: at least one heart.
Answer
4 cards can be drawn out of 52 cards in ${ }^{52} \mathrm{C}_4$ ways.
$
\therefore \mathrm{n}(\mathrm{S})={ }^{52} \mathrm{C}_4
$
Let $\mathrm{C}$ be the event that out of the four cards drawn at least one is a heart.
$\therefore C^{\prime}$ is the event that all 4 cards drawn are non-heart cards.
In a pack of 52 cards, there are 39 non-heart cards.
$\therefore 4$ non-heart cards can be drawn in ${ }^{39} \mathrm{C}_4$ ways.
$
\begin{aligned}
& \therefore \mathrm{n}\left(\mathrm{C}^{\prime}\right)={ }^{39} \mathrm{C}_4 \\
& \therefore \mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{\mathrm{n}\left(\mathrm{C}^7\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{39} \mathrm{C}_4}{{ }^{52} \mathrm{C}_4} \\
& \therefore \mathrm{P}(\mathrm{C})=1-\mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\frac{{ }^{99} \mathrm{C}_4}{{ }^{52} \mathrm{C}_4}
\end{aligned}
$
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Question 252 Marks
Four cards are drawn from a pack of 52 cards. Find the probability that:all the cards are from different suits.
Answer
4 cards can be drawn out of 52 cards in ${ }^{52} \mathrm{C}_4$ ways.
$
\therefore \mathrm{n}(\mathrm{S})={ }^{52} \mathrm{C}_4
$
Let $\mathrm{B}$ be the event that all the cards drawn are of different suits.
A pack of 52 cards consists of 4 suits each containing 13 cards.
$\therefore$ A card can be drawn from each suit in ${ }^{13} \mathrm{C}_1$ ways.
$\therefore 4$ cards can be drawn from 4 different suits in ${ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1$ ways.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{B})={ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1}{{ }^{52} \mathrm{C}_4}
\end{aligned}
$
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Question 262 Marks
Four cards are drawn from a pack of 52 cards. Find the probability that: 3 are Kings and 1 is Jack.
Answer
4 cards can be drawn out of 52 cards in ${ }^{52} \mathrm{C}_4$ ways.
$
\therefore \mathrm{n}(\mathrm{S})={ }^{52} \mathrm{C}_4
$
Let $\mathrm{A}$ be the event that out of the four cards drawn, 3 are kings and 1 is a jack. There are 4 kings and 4 jacks in a pack of 52 cards.
$\therefore 3$ kings can be drawn from 4 kings in ${ }^4 C_3$ ways.
Similarly, 1 jack can be drawn out of 4 jacks in ${ }^4 C_1$ ways.
$\therefore$ Total number of ways in which 3 kings and 1 jack can be drawn is ${ }^4 C_3 \times{ }^4 C_1$
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})={ }^4 \mathrm{C}_3 \times{ }^4 \mathrm{C}_1 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^4 \mathrm{C}_3 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_4}
\end{aligned}
$
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Question 272 Marks
Two cards are drawn from a pack of 52 cards. Find the probability that: both are from the same denomination.
Answer
Two cards can be drawn from a pack of 52 cards in ${ }^{52} \mathrm{C}_2$ ways.
$
\therefore \mathrm{n}(5)={ }^{52} \mathrm{C}_2
$
Let $G$ be the event that both the cards drawn are of same denominations.
A pack of cards has 13 denominations and 4 different cards for each denomination
$
\begin{aligned}
& \therefore n(G)=13 \times{ }^4 C_2 \\
& \therefore P(G)=\frac{n(G)}{n(B)}=\frac{13 \times{ }^4 C_2}{{ }^2 C_2}
\end{aligned}
$
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Question 282 Marks
Two cards are drawn from a pack of 52 cards. Find the probability that:both are from the same suit.
Answer
Two cards can be drawn from a pack of 52 cards in ${ }^{52} \mathrm{C}_2$ ways.
$
\therefore \mathrm{n}(5)={ }^{52} \mathrm{C}_2
$
Let $\mathrm{E}$ be the event that out of the two cards drawn one is a spade and other is non-spade. There are 13 spade cards and 39 cards are non-spade cards in a pack of 52 cards.
$\therefore$ One spade card can be drawn from 13 spade cards in ${ }^{13} C_1$ ways and one non-spade card can be drawn from 39 non-spade cards in $39 \mathrm{C} 1$ ways.
$
\begin{aligned}
& \therefore n(E)={ }^{13} C_1 \cdot{ }^{39} C_1 \\
& \therefore P(E)=\frac{n(K)}{n(S)}=\frac{{ }^{13} C_1 \cdot{ }^{27} C_1}{{ }^2 C_2}
\end{aligned}
$
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Question 292 Marks
Two cards are drawn from a pack of 52 cards. Find the probability that: one is a spade and the other is a non-spade
Answer
Two cards can be drawn from a pack of 52 cards in ${ }^{52} \mathrm{C}_2$ ways.
$
\therefore \mathrm{n}(5)={ }^{52} \mathrm{C}_2
$
Let $\mathrm{D}$ be the event that both the cards drawn are face cards.
There are 12 face cards in a pack of 52 cards.
$\therefore 2$ face cards can be drawn from 12 face cards in ${ }^{12} C_2$ ways.
$\therefore \mathrm{n}(\mathrm{D})={ }^{12} \mathrm{C}_2$
$
\therefore \mathrm{P}(\mathrm{D})=\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{B})}=\frac{{ }^{12} \mathrm{C}_2}{{ }^{3 \mathrm{~S}} \mathrm{C}_2}
$
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Question 302 Marks
Two cards are drawn from a pack of 52 cards. Find the probability that:both are face cards.
Answer
Two cards can be drawn from a pack of 52 cards in ${ }^{52} \mathrm{C}_2$ ways.
$
\therefore \mathrm{n}(5)={ }^{52} \mathrm{C}_2
$
Let $\mathrm{D}$ be the event that both the cards drawn are face cards.
There are 12 face cards in a pack of 52 cards.
$\therefore 2$ face cards can be drawn from 12 face cards in ${ }^{12} C_2$ ways.
$\therefore \mathrm{n}(\mathrm{D})={ }^{12} \mathrm{C}_2$
$
\therefore \mathrm{P}(\mathrm{D})=\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{B})}=\frac{{ }^{12} \mathrm{C}_2}{{ }^{3 \mathrm{~S}} \mathrm{C}_2}
$
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Question 312 Marks
Two cards are drawn from a pack of 52 cards. Find the probability that:both are ace cards.
Answer
Two cards can be drawn from a pack of 52 cards in ${ }^{52} \mathrm{C}_2$ ways.
$
\therefore \mathrm{n}(5)={ }^{52} \mathrm{C}_2
$
Let $C$ be the event that both the cards drawn are aces.
In a pack of 52 cards, there are 4 ace cards.
$\therefore 2$ ace cards can be drawn from 4 ace cards in ${ }^4 C_2$ ways
$
\begin{aligned}
\therefore \mathrm{n}(\mathrm{C}) & ={ }^4 \mathrm{C}_2 \\
\therefore \mathrm{P}(\mathrm{C}) & =\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{s})}=\frac{{ }^4 \mathrm{C}_2}{{ }^5 \mathrm{C}_2}
\end{aligned}
$
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Question 322 Marks
Two cards are drawn from a pack of 52 cards. Find the probability that: both are diamonds.
Answer
Two cards can be drawn from a pack of 52 cards in ${ }^{52} \mathrm{C}_2$ ways.
$
\therefore \mathrm{n}(5)={ }^{52} \mathrm{C}_2
$
Let $B$ be the event that both the cards drawn are diamond.
There are 13 diamond cards in a pack of 52 cards.
$\therefore 2$ diamond cards can be drawn from 13 diamond cards in ${ }^{13} C_2$ ways
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{B})={ }^{13} \mathrm{C}_2 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{B})}=\frac{{ }^{13} \mathrm{C}_1}{{ }^{{ }^3 \mathrm{C}_1}}
\end{aligned}
$
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Question 332 Marks
Two cards are drawn from a pack of 52 cards. Find the probability that: both are black.
Answer
Two cards can be drawn from a pack of 52 cards in ${ }^{52} \mathrm{C}_2$ ways.
$
\therefore \mathrm{n}(5)={ }^{52} \mathrm{C}_2
$
Let $A$ be the event that both the cards drawn are black.
The pack of 52 cards contains 26 black cards.
$\therefore 2$ cards can be drawn from them in ${ }^{26} \mathrm{C}_2$ ways
$
\begin{aligned}
& \therefore n(A)={ }^{2 f_1} C_2 \\
& \therefore P(A)=\frac{n(A)}{n(B)}=\frac{{ }^{3 s} C_2}{{ }^3 C_2}
\end{aligned}
$
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