Solve the following Question.(1 Marks) - Maths (commerce) STD 11 Commerce / Arts Questions - Vidyadip

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Solve the following Question.(1 Marks)

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

Question 11 Mark

From a group of 2 men $\left(M_1, M_2\right)$ and three women $\left(W_1, W_2, W_3\right)$, two persons are selected. Describe the sample space of the experiment. If $E$ is the event in which one man and one woman are selected, then which are the cases favourable to $E$ ?

Answer

Let $S$ be the sample space of the given event.
$
\begin{aligned}
& \therefore \mathrm{S}=\left\{\left(\mathrm{M}_1, \mathrm{M}_2\right),\left(\mathrm{M}_1, \mathrm{~W}_1\right),\left(\mathrm{M}_1, \mathrm{~W}_2\right),\left(\mathrm{M}_1, \mathrm{~W}_3\right),\left(\mathrm{M}_2, \mathrm{~W}_1\right),\left(\mathrm{M}_2, \mathrm{~W}_2\right),\left(\mathrm{M}_2, \mathrm{~W}_3\right),\left(\mathrm{W}_1, \mathrm{~W}_2\right)\left(\mathrm{W}_1, \mathrm{~W}_3\right),\right. \\
& \left.\left(\mathrm{W}_2, \mathrm{~W}_3\right)\right\}
\end{aligned}
$
Let $E$ be the event that one man and one woman are selected.
$
\therefore E=\left\{\left(M_1, W_1\right),\left(M_1, W_2\right),\left(M_1, W_3\right),\left(M_2, W_1\right),\left(M_2, W_2\right),\left(M_2, W_3\right)\right\}
$
Here, the order is not important in which 2 persons are selected e.g. $\left(M_1, M_2\right)$ is the same as $\left(\mathrm{M}_2, \mathrm{M}_1\right)$

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Question 21 Mark

Two cards are drawn one after the other from a pack of 52 cards with replacement. What is the probability that both the cards are drawn are face cards?

Answer

Two cards are drawn from a pack of 52 cards with replacement.
$
\therefore \mathrm{n}(\mathrm{S})=52 \times 52
$
Let $\mathrm{A}$ be the event that two cards drawn are face cards.
First card from 12 face cards is drawn with replacement in ${ }^{12} \mathrm{C}_1=12$ ways and second face card is drawn from 12 face card in ${ }^{12} C_1=12$ ways after replacement.
$
\therefore \mathrm{n}(\mathrm{A})=12 \times 12
$
$\therefore \mathrm{P}($ that both the cards drawn are face cards $)=\mathrm{P}(\mathrm{A})$
$
=\frac{n(A)}{n(S)}=\frac{12 \times 12}{52 \times 52}=\frac{9}{169}
$

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Question 31 Mark

If $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$. Find the values of the following probabilities.: $P\left(A^{\prime} \cap B^{\prime}\right)$

Answer

Here, $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$
$
\begin{aligned}:P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime} \ldots . . .[\text { De Morgan's law }] \\
& =1-P(A \cup B) \\
& =1-\frac{1}{2} \\
& =\frac{1}{2}
\end{aligned}
$

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Question 41 Mark

If $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$. Find the values of the following probabilities.: $P\left(A^{\prime} \cup B^{\prime}\right)$

Answer

Here, $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$
$
\begin{aligned}:P\left(A^{\prime} \cup B^{\prime}\right)=P(A \cap B)^{\prime} \ldots . .[\text { [De Morgan's law] } \\
& =1-P(A \cap B) \\
& =1-\frac{3}{20} \\
& =\frac{17}{20}
\end{aligned}
$
$
\begin{aligned}

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Question 51 Mark

If $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$. Find the values of the following probabilities.: $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$

Answer

Here, $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$
$
\begin{aligned}:P\left(A^{\prime} \cap B\right)=P(B)-P(A \cap B) \\
& =\frac{2}{5}-\frac{3}{20} \\
& =\frac{1}{4}
\end{aligned}
$
$
\begin{aligned}

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Question 61 Mark

If $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$. Find the values of the following probabilities.: $P\left(A \cap B^{\prime}\right)$

Answer

Here, $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$
$
\begin{aligned}:P\left(A^{\prime} \cap B^{\prime}\right)=P(A)-P(A \cap B) \\
& =\frac{1}{4}-\frac{3}{20} \\
& =\frac{1}{10}
\end{aligned}
$
$
\begin{aligned}

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Question 71 Mark

If $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$. Find the values of the following probabilities.: $P(A \cap B)$

Answer

Here, $P(A)=\frac{1}{4}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{1}{2}$
$
\begin{aligned}: P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \therefore P(A \cap B)=P(A)+P(B)-P(A \cup B) \\
& =\frac{1}{4}+\frac{2}{5}-\frac{1}{2} \\
& =\frac{3}{20}
\end{aligned}
$
$
\begin{aligned}

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Question 81 Mark

From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains: at least 3 boys.

Answer

The group consists of 8 boys and 5 girls i.e., $8+5=13$ persons.
A committee of 5 is to be formed from this group.
$\therefore 5$ persons from 13 persons can be selected in ${ }^{13} \mathrm{C}_5$ ways
$\therefore \mathrm{n}(\mathrm{S})={ }^{13} \mathrm{C}_5$

Let $\mathrm{B}$ be the event that the committee contains at least 3 boys (i.e, 3 boys and 2 girls or 4 boys and 1 girl or 5 boys and no girl)
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{B})={ }^8 \mathrm{C}_3 \cdot{ }^5 \mathrm{C}_2+{ }^8 \mathrm{C}_4 \cdot{ }^5 \mathrm{C}_1+{ }^8 \mathrm{C}_5 \cdot{ }^5 \mathrm{C}_0 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^8 \mathrm{C}_3{ }^5 \mathrm{C}_2+{ }^8 \mathrm{C}_4 \cdot{ }^5 \mathrm{C}_1+{ }^8 \mathrm{C}_5{ }^5 \mathrm{C}_0}{{ }^{13} \mathrm{C}_5}
\end{aligned}
$

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Question 91 Mark

From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains: 3 boys and 2 girls

Answer

The group consists of 8 boys and 5 girls i.e., $8+5=13$ persons.
A committee of 5 is to be formed from this group.
$\therefore 5$ persons from 13 persons can be selected in ${ }^{13} \mathrm{C}_5$ ways
$\therefore \mathrm{n}(\mathrm{S})={ }^{13} \mathrm{C}_5$

Let $\mathrm{A}$ be the event that the committee contains 3 boys and 2 girls.
3 boys from 8 boys can be selected in ${ }^8 \mathrm{C}_3$ ways and 2 girls from 5 girls can be selected in ${ }^5 \mathrm{C}_2$
ways
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})={ }^8 \mathrm{C}_3 \cdot{ }^5 \mathrm{C}_2 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^8 \mathrm{C}_3{ }^5 \mathrm{C}_2}{{ }^{13} \mathrm{C}_5}
\end{aligned}
$

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Question 101 Mark

A box contains 75 tickets numbered 1 to 75 . A ticket is drawn at random from the box. What is the probability that the: number on the ticket is divisible by 3 and 5 ?

Answer

The box contains 75 tickets numbered 1 to 75 .
$\therefore 1$ ticket can be drawn from the box in ${ }^{75} \mathrm{C}_1=75$ ways.
$\therefore \mathrm{n}(\mathrm{S})=75$: Let $\mathrm{D}$ be the event that number on ticket is divisible by 3 and 5 i.e., divisible by L.C.M. of 3 and 5 i.e., 15
$
\begin{aligned}
& \therefore D=\{15,30,45,60,75\} \\
& \therefore \mathrm{n}(\mathrm{D})=5 \\
& \therefore P(D)=\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{5}{75}=\frac{1}{15}
\end{aligned}
$

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Question 111 Mark

A box contains 75 tickets numbered 1 to 75 . A ticket is drawn at random from the box. What is the probability that the: number on the ticket is prime?

Answer

The box contains 75 tickets numbered 1 to 75 .
$\therefore 1$ ticket can be drawn from the box in ${ }^{75} \mathrm{C}_1=75$ ways.
$\therefore \mathrm{n}(\mathrm{S})=75$: Let $\mathrm{C}$ be the event that the number on the ticket is a prime number.
$
\begin{aligned}
& \therefore \mathrm{C}=\{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73\} \\
& \therefore \mathrm{n}(\mathrm{C})=21 \\
& \therefore \mathrm{P}(\mathrm{C})=\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{s})}=\frac{21}{75}=\frac{7}{25}
\end{aligned}
$

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Question 121 Mark

A box contains 75 tickets numbered 1 to 75 . A ticket is drawn at random from the box. What is the probability that the: number on the ticket is a perfect square?

Answer

The box contains 75 tickets numbered 1 to 75 .
$\therefore 1$ ticket can be drawn from the box in ${ }^{75} \mathrm{C}_1=75$ ways.
$\therefore \mathrm{n}(\mathrm{S})=75$: Let $\mathrm{B}$ be the event that number on ticket is a perfect square.
$
\begin{aligned}
& \therefore \mathrm{B}=\{1,4,9,16,25,36,49,64\} \\
& \therefore \mathrm{n}(\mathrm{B})=8 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{8}{75}
\end{aligned}
$

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Question 131 Mark

A box contains 75 tickets numbered 1 to 75 . A ticket is drawn at random from the box. What is the probability that the: number on the ticket is divisible by 6 ?

Answer

The box contains 75 tickets numbered 1 to 75 .
$\therefore 1$ ticket can be drawn from the box in ${ }^{75} \mathrm{C}_1=75$ ways.
$\therefore \mathrm{n}(\mathrm{S})=75$: Let $\mathrm{A}$ be the event that number on ticket is divisible by 6 .
$
\begin{aligned}
& \therefore A=\{6,12,18,24,30,36,42,48,54,60,66,72\} \\
& \therefore \mathrm{n}(\mathrm{A})=12 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{12}{75}=\frac{4}{25}
\end{aligned}
$

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Question 141 Mark

A fair die is thrown two times. Find the chance that: both the times die to show the same number (doublet).

Answer

If a fair die is thrown twice, the sample space is
$
S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,
$
$3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
$\therefore \mathrm{n}(\mathrm{S})=36$: Let $\mathrm{F}$ be the event that both times die shows same number.
$
\begin{aligned}
& \therefore \mathrm{F}=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\} \\
& \therefore \mathrm{n}(\mathrm{F})=6 \\
& \therefore \mathrm{P}(\mathrm{F})=\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}
\end{aligned}
$

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Question 151 Mark

A fair die is thrown two times. Find the chance that:the first throw gives an odd number and the second throw gives a multiple of 3.

Answer

Let $E$ be the event that 1st throw gives an odd number and 2 nd throw gives multiple of 3 . $ \begin{aligned} & \therefore \mathrm{E}=\{(1,3),(1,6),(3,3),(3,6),(5,3),(5,6)\} \\ & \therefore \mathrm{n}(\mathrm{E})=6 \\ & \therefore \mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6} \end{aligned} $

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Question 161 Mark

A fair die is thrown two times. Find the chance that: sum of the numbers on the upper face is 4.

Answer

If a fair die is thrown twice, the sample space is
$
S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,
$
$3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
$\therefore \mathrm{n}(\mathrm{S})=36$: Let $\mathrm{D}$ be the event that sum of the numbers on uppermost face is 4 .
$
\begin{aligned}
& \therefore D=\{(1,3),(2,2),(3,1)\} \\
& \therefore \mathrm{n}(\mathrm{D})=3 \\
& \therefore \mathrm{P}(\mathrm{D})=\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}=\frac{1}{12}
\end{aligned}
$

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Question 171 Mark

A fair die is thrown two times. Find the chance that:sum of the numbers on the upper face is at least 10.

Answer

If a fair die is thrown twice, the sample space is
$
S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,
$
$3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
$\therefore \mathrm{n}(\mathrm{S})=36$:Let $\mathrm{C}$ be the event that sum of the numbers on uppermost face is at least 10 (i.e., 10 or more than 10 which are 10 or 11 or 12)
$
\begin{aligned}
& \therefore C=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\} \\
& \therefore \mathrm{n}(\mathrm{C})=6 \\
& \therefore \mathrm{P}(\mathrm{C})=\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}
\end{aligned}
$

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Question 181 Mark

A fair die is thrown two times. Find the chance that: sum of the numbers on the upper face is 10.

Answer

If a fair die is thrown twice, the sample space is
$
S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,
$
$3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
$\therefore \mathrm{n}(\mathrm{S})=36$: Let $B$ be the event that sum of the numbers on uppermost face is 10 .
$
\begin{aligned}
& \therefore \mathrm{B}=\{(4,6),(5,5),(6,4)\} \\
& \therefore \mathrm{n}(\mathrm{B})=3 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}
\end{aligned}
$

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Question 191 Mark

A fair die is thrown two times. Find the chance that: product of the numbers on the upper face is 12.

Answer

If a fair die is thrown twice, the sample space is
$
S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,
$
$3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
$\therefore \mathrm{n}(\mathrm{S})=36$: Let $\mathrm{A}$ be the event that the product of the numbers on uppermost face is 12 .
$
\begin{aligned}
& \therefore A=\{(2,6),(3,4),(4,3),(6,2)\} \\
& \therefore \mathrm{n}(\mathrm{A})=4 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{4}{36}=\frac{1}{9}
\end{aligned}
$

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Question 201 Mark

Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample space, if cards are drawn: without replacement.

Answer

The bag contains 4 cards marked 5, 6, 7, and 8 .
Two cards are to be drawn from this bag.: If the two cards are drawn without replacement, then the sample space is $S=\{(5,6),(5,7),(5,8),(6,5),(6,7),(6,8),(7,5),(7,6),(7,8),(8,5),(8,6),(8,7)\}$

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Question 211 Mark

Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample space, if cards are drawn: with replacement

Answer

The bag contains 4 cards marked 5, 6, 7, and 8 .
Two cards are to be drawn from this bag.: If the two cards are drawn with replacement, then the sample space is $S=\{(5,5),(5,6),(5,7),(5,8),(6,5),(6,6),(6,7),(6,8),(7,5),(7,6),(7,7),(7,8),(8,5),(8,6),(8$,
7), $(8,8)\}$

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Question 221 Mark

Box-I contains 3 red (R11, R12, R13) and 2 blue (B11, B12) marbles while Box-II contains 2 red (R21, R22) and 4 blue (B21, B22, B23, B24) marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box-l; if it turns up tails, a marble is chosen from Box-II.
Describe the sample space.

Answer

Box I contains 3 red and 2 blue marbles i.e., (R11, R12, R13, B11, B12)
Box Il contains 2 red and 4 blue marbles i.e., (R21, R22, B21, B22, B23, B24)
It is given that a fair coin is tossed and if a head comes then marble is chosen from box I otherwise it is chosen from box II
$\therefore$ the sample space is
$S=\{(H, R 11),(H, R 12),(H, R 13),(H, B 11),(H, B 12),(T, R 21),(T, R 22),(T, B 21),(T, B 22),(T, B 23)$,
$(\mathrm{T}, \mathrm{B} 24)\}$

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Question 231 Mark

A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space, if
consideration of suits: is taken into account.

Answer

If consideration of suits is taken into account, then one card can be drawn from each suit
$
\begin{aligned}
& \text { in }{ }^{13} C_1 \times{ }^4 C_1 \\
& =13 \times 4 \\
& =52 \text { ways. } \\
& \therefore \mathrm{n}(\mathrm{S})=52
\end{aligned}
$
Question 7.
Box-I contains 3 red (R11, R12, R13) and 2 blue (B11, B12) marbles while Box-II contains 2 red (R21, R22) and 4 blue (B21, B22, B23, B24) marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box-l; if it turns up tails, a marble is chosen from Box-II.
Describe the sample space.
Solution:
Box I contains 3 red and 2 blue marbles i.e., (R11, R12, R13, B11, B12)
Box Il contains 2 red and 4 blue marbles i.e., (R21, R22, B21, B22, B23, B24)
It is given that a fair coin is tossed and if a head comes then marble is chosen from box I otherwise it is chosen from box II
$\therefore$ the sample space is
$S=\{(H, R 11),(H, R 12),(H, R 13),(H, B 11),(H, B 12),(T, R 21),(T, R 22),(T, B 21),(T, B 22),(T, B 23)$,
$(T, B 24)\}$

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Question 241 Mark

A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space, if
consideration of suits: is not taken into account.

Answer

If consideration of suits is not taken into account, then one card can be drawn from the pack of 52 playing cards in ${ }^{52} \mathrm{C}_1=52$ ways.
$
\therefore \mathrm{n}(\mathrm{S})=52
$

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Question 251 Mark

Find the total number of distinct possible outcomes $n(S)$ for each of the following random experiments.: 6 students are arranged in a row for photographs.

Answer

Six students can be arranged in a row for a photograph in ${ }^6 P_6=6$ ! ways.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{S})=6 ! \\
& =6 \times 5 \times 4 \times 3 \times 2 \times 1 \\
& =720
\end{aligned}
$

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Question 261 Mark

Find the total number of distinct possible outcomes $n(S)$ for each of the following random experiments.: 5 balls are randomly placed into five cells, such that each cell will be occupied.

Answer

5 balls have to be placed in 5 cells in such a way that each cell is occupied.
$\therefore$ The first ball can be placed into one of the 5 cells in 5 ways, the second ball placed into one of the remaining 4 cells in 4 ways.
Similarly, the third, fourth, and fifth balls can be placed in 3 ways, 2 ways, and 1 way, respectively.
$
\begin{aligned}
& \therefore \text { a total number of ways of filling } 5 \text { cells such that each cell is occupied }=5 ! \\
& =5 \times 4 \times 3 \times 2 \times 1 \\
& =120 \\
& \therefore \mathrm{n}(\mathrm{S})=120
\end{aligned}
$

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Question 271 Mark

Find the total number of distinct possible outcomes $n(S)$ for each of the following random experiments.: From a group of 4 boys and 3 girls, any two students are selected at random.

Answer

There are 4 boys and 3 girls i.e., 7 students.
2 students can be selected from these 7 students in ${ }^7 \mathrm{C}_2$ ways.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{S})={ }^7 \mathrm{C}_2 \\
& =\frac{7 \times 6}{2 \times 1} \\
& =21
\end{aligned}
$

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Question 281 Mark

Find the total number of distinct possible outcomes $n(S)$ for each of the following random experiments.: From a box containing 25 lottery tickets and 3 tickets are drawn at random.

Answer

Let $\mathrm{S}$ be the event that 3 tickets are drawn at random from 25 tickets
$\therefore 3$ tickets can be selected in ${ }^{25} \mathrm{C}_3$ ways
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{S})={ }^{25} \mathrm{C}_3 \\
& =\frac{25 \times 24 \times 23}{3 \times 2 \times 1} \\
& =2300
\end{aligned}
$

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Question 291 Mark

A coin and die are tossed. State sample space of following events.: C: Getting a tail and perfect square.

Answer

When a coin and a die are tossed the sample space $\mathrm{S}$ is given by $S=\{H 1, H 2, H 3, H 4, H 5, H 6, T 1, T 2, T 3, T 4, T 5, T 6\}$: $\mathrm{C}$ : getting a tail and a perfect square.
$
\therefore \mathrm{C}=\{\mathrm{T} 1, \mathrm{~T} 4\}
$

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Question 301 Mark

A coin and die are tossed. State sample space of following events.: $\mathrm{B}$ : Getting a prime number.

Answer

When a coin and a die are tossed the sample space $\mathrm{S}$ is given by $S=\{H 1, H 2, H 3, H 4, H 5, H 6, T 1, T 2, T 3, T 4, T 5, T 6\}$: $\mathrm{B}$ : getting a prime number
$
\therefore \mathrm{B}=\{\mathrm{H} 2, \mathrm{H} 3, \mathrm{H} 5, \mathrm{~T} 2, \mathrm{~T} 3, \mathrm{~T} 5\}
$

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Question 311 Mark

A coin and die are tossed. State sample space of following events.: A: Getting a head and an even number.

Answer

When a coin and a die are tossed the sample space $\mathrm{S}$ is given by $S=\{H 1, H 2, H 3, H 4, H 5, H 6, T 1, T 2, T 3, T 4, T 5, T 6\}$: A: getting a head and an even number
$
\therefore \mathrm{A}=\{\mathrm{H} 2, \mathrm{H} 4, \mathrm{H} 6\}
$

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Question 321 Mark

In a bag, there are three balls; one black, one red, and one green. Two balls are drawn one after another with replacement. State sample space and n(S).

Answer

The bag contains 3 balls out of which one is black $(B)$, one is red $(R)$ and the other one is green $(G)$.
Two balls are drawn one after the other, with replacement, from the bag. $\therefore$ the sample space $S$ is given by
$
\begin{aligned}
& \mathrm{S}=\{\mathrm{BB}, \mathrm{BR}, \mathrm{BG}, \mathrm{RB}, \mathrm{RR}, \mathrm{RG}, \mathrm{GB}, \mathrm{GR}, \mathrm{GG}\} \\
& \therefore \mathrm{n}(\mathrm{S})=9
\end{aligned}
$

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Question 331 Mark

A coin is tossed twice. If a second throw results in a head, a die is thrown, otherwise, a coin is tossed.

Answer

When a coin is tossed twice, the outcomes are $\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}$.
Let $\mathrm{A}$ be the event that the second throw results in the head when a coin is tossed twice followed by a die is thrown.
$\therefore \mathrm{A}=\{\mathrm{HH} 1, \mathrm{HH} 2, \mathrm{HH} 3, \mathrm{HH} 4, \mathrm{HH} 5, \mathrm{HH} 6, \mathrm{TH} 1, \mathrm{TH} 2, \mathrm{TH} 3, \mathrm{TH} 4, \mathrm{TH} 5, \mathrm{TH} 6\}$
The remaining outcomes i.e., HT, TT are followed by the tossing of a coin.
Let us consider this as event $\mathrm{B}$.
$\therefore \mathrm{B}=\{\mathrm{HTT}, \mathrm{HTH}, \mathrm{TTT}, \mathrm{TTH}\}$
The sample space $\mathrm{S}$ of the experiment is $\mathrm{A} \cup \mathrm{B}$.
$\therefore \mathrm{S}=\{\mathrm{HH} 1, \mathrm{HH} 2, \mathrm{HH} 3, \mathrm{HH} 4, \mathrm{HH} 5, \mathrm{HH} 6, \mathrm{TH} 1, \mathrm{TH} 2, \mathrm{TH} 3, \mathrm{TH} 4, \mathrm{TH} 5, \mathrm{TH} 6, \mathrm{HTT}$,
$\mathrm{HTH}, \mathrm{TTT}, \mathrm{TTH}\}$
$
\therefore \mathrm{n}(\mathrm{S})=16
$

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Question 341 Mark

State the sample space and $\mathrm{n}(\mathrm{S})$ for the following random experiments.: A coin is tossed twice. If a second throw results in a tail, a die is thrown.

Answer

When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
A coin is tossed twice and if the second throw results in a tail, a die is thrown. Out of the above 4 possibilities, on second throw tail occurs in two cases only i.e., HT, TT.
$\therefore \mathrm{S}=\{\mathrm{HH}, \mathrm{TH}, \mathrm{HT} 1, \mathrm{HT} 2, \mathrm{HT} 3, \mathrm{HT} 4, \mathrm{HT} 5, \mathrm{HT} 6, \mathrm{TT} 1, \mathrm{TT} 2, \mathrm{TT} 3, \mathrm{TT} 4, \mathrm{TT} 5, \mathrm{TT} 6\}$
$\therefore \mathrm{n}(\mathrm{S})=14$

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