2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip

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Question 12 Marks

There are three events A, B, C one of which must and only one can happen, the odds are 8 to 3 against A, 5 to 2 against B, find the odds against C.

Answer

Given,
$\text{P}(\overline{\text{A}}):\text{P}(\text{B})=8:3$
$\Rightarrow\frac{1-\text{P}(\text{A})}{\text{P}(\text{A})}=\frac{8}{3}$
$\Rightarrow\text{P}(\text{A})=\frac{3}{11}$
$\text{P}(\overline{\text{B}}):\text{P}(\text{B})=5:2$
$\Rightarrow\frac{1-\text{P}(\text{B})}{\text{P}(\text{B})}=\frac{5}{2}$
$\Rightarrow\frac{1}{\text{P}(\text{B})}=\frac{5}{2}+1=\frac{7}{2}$
$\Rightarrow\text{P}(\text{B})=\frac{2}{7}$
$\because$ A, B and C are mutually exhaustive
$\therefore\text{A}\cup\text{B}\cup\text{C}=\text{S}$
$\Rightarrow\text{P}(\text{A}\cup\text{B}\cup\text{C})=\text{P}(\text{S})$
$\Rightarrow\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{C})=1$
$\text{P}(\text{C})=1-\big\{\text{P}(\text{A})+\text{P}(\text{B})\big\}$
$=1-\Big(\frac{3}{11}+\frac{2}{7}\Big)$
$=1-\frac{43}{77}$
$=\frac{34}{77}$
$\Rightarrow\text{P}(\overline{\text{C}})=1-\text{P}(\text{C})$
$=1-\frac{34}{77}$
$=\frac{43}{77}$
$\therefore$ Odds against C is
$\text{P}(\overline{\text{C}}):\text{P}(\text{C})=\frac{43}{77}:\frac{34}{77}$
$=43:34$

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Question 22 Marks

In a simultanepous throw of a pair of dice, find the probability of getting,A total of 9 or 11

Answer

Let E be the events that a total of 9 or 11 appear on the faces of dice.
$\therefore\ \text{E}=\big\{(3, 6),\ (4, 5),\ (5, 4),\ (5, 6),\ (6, 3),\ (6,5)\big\}$
$\Rightarrow\text{n(E)}=6$
$\therefore\text{P(E)}=\frac{6}{36}=\frac{1}{6}$

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Question 32 Marks

A and B throw a pair of dice. If A throws 9, find B's chance of throwing a higher number.

Answer

$\because$ A and B throw a pair of dice
$\therefore\text{n}(\text{S})=6^2=36$
Let E be the event that A throw 9 and B throw more than 9, that is 10, 11, 12
$\therefore\big\{(4,\ 6),\ (5,\ 5),\ (5,\ 6),\ (6,\ 4),\ (6,\ 5),\ (6,\ 6)\big\}$
$\therefore\text{n}(\text{S})=6$
$\therefore\text{p}(\text{E})=\frac{6}{36}=\frac{1}{6}$
$\therefore\text{p}(\text{E})=\frac{1}{6}$

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Question 42 Marks

A bag contains $8$ red, $3$ white and $9$ blue balls. if three balls are drawn at random, determine the probability that:

All the three balls are blue balls
All the balls are of different colours.

Answer

Bag: $8 -$ Red, $3 -$ White, $9 -$ Blue,
since three balls are drawn
$\therefore\text{n}\text{(s)}=\ ^{20}\text{C}_3$

Let $E$ be the event that all the three balls are blue

$\therefore\text{n}\text{(E)}=\ ^{9}\text{C}_3$
$\therefore\text{P}\text{(s)}=\frac{\ ^{9}\text{C}_3}{\ ^{20}\text{C}_3}$
$=\frac{9\times8\times7}{20\times19\times18}=\frac{7}{95}$

Let $E$ be the event that all the balls are of different colour.

$\therefore\text{n}\text{(s)}=\ ^{8}\text{C}_1\times\ ^{3}\text{C}_1\times\ ^{9}\text{C}_1$
$\therefore\text{p}\text{(E)}=\frac{\ ^{8}\text{C}_1\times\ ^{3}\text{C}_1\times\ ^{9}\text{C}_1}{\ ^{20}\text{C}_3}$
$=\frac{18}{95}$

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Question 52 Marks

In a single throw of three dice, find the probability of getting the same number on all the three dice.

Answer

Three dice are rolled then,
$\text{n}\text{(S)} = 6^3 ,= 216$
E be the event of getting same numbers on all the three dice
$\text{E} = \big\{\text{(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)}\big\}$
$\therefore\text{n(E)=6}$
$\therefore\text{p(E)}=\frac{6}{216}=\frac{1}{36}$
$\text{p(E)}=\frac{1}{36}$

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Question 62 Marks

In a simultanepous throw of a pair of dice, find the probability of getting,A sum less than 6.

Answer

Let E be the events that less than 6 as a sum offer on the faces of dice.$\therefore\ \text{E}=\big\{(1, 1),\ (1, 2),\ (1, 3),\ (1, 4),\ (2, 1),\ (2, 2),\ (2, 3),\ (3, 1),\ (3, 2),\ (4, 1)\big\}$
$\therefore\ \text{E}=10$ $\therefore\ \text{p(E)}=\frac{10}{36}=\frac{5}{18}$

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Question 72 Marks

State true or false:

$A$ and $B$ are mutually exclusive.
$A$ and $B$ are mutually exclusive and exhaustive events.
$A$ and $C$ are mutually exclusive events.
$C$ and $D$ are mutually exclusive and exhaustive events.
$C, D$ and $E$ are mutually exclusive and exhaustive events.
$A'$ and $B'$ are mutually exclusive events.
$A, B, F$ are mutually exclusive and exhaustive events.

Answer

True, because $A ∩ B = \Phi$
True, because $A ∩ B = \Phi$ and $A ∪ B = S$
False, because $A ∩ C \neq \Phi$
False, because $C ∩ D = \Phi$ but $C ∪ D \neq S$
True, because $C ∩ D ∩ E = \Phi$ and $C ∪ D ∪ E = S$
True, because $A' ∩ B' = \Phi$
False, because $A ∩ B ∩ F \neq \Phi$ and $A ∪ B ∪ F = S$

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Question 82 Marks

A bag contains $7$ white, $5$ black and $4$ red balls. If two balls are drawn at random, find the probability that:

Both the balls are white
One ball is black and the other red
Both the balls are of the same colour.

Answer

$\text{BAG 7-}$white ball
$5-$black ball
$4-$blue ball
$\because$ Two balls are drawn
$\therefore\text{n}(\text{S})=\ ^{16}\text{C}_2$

Let $E$ be the event that both the balls are white

$\therefore\text{n}(\text{E})=\ ^{7}\text{C}_2$
$\therefore\text{p}(\text{E})=\ \frac{^{7}\text{C}_2}{^{16}\text{C}_2}=\frac{7\times6}{16\times15}=\frac{7}{40}$
$\therefore\text{p}(\text{E})=\frac{7}{40}$

Let $E$ be the event that, one ball and one red ball is drawn

$\therefore\text{n}(\text{E})=\ ^5\text{C}_1\times^4\text{C}_1$
$\therefore\text{p}(\text{E})=\frac{^{5}\text{C}_1\times^{4}\text{C}_1}{^{16}\text{C}_1}=\frac{5\times4\times2}{16\times15}=\frac{1}{6}$
$\therefore\text{p}(\text{E})=\frac{1}{6}$

Let $E$ be the event that both the balls are of the same colour.

$\therefore\text{n}(\text{E})=\ ^7\text{C}_2$ or $^{5}\text{C}_2$ or $^4\text{C}_2$
$\therefore\text{p}(\text{E})=\ \frac{^{7}\text{C}_2+^{5}\text{C}_2+^4\text{C}_2}{^{16}\text{C}_2}$
$=\frac{7\times6+5\times4+4\times2}{16\times15}$
$=\frac{70}{240}$
$=\frac{7}{24}$

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Question 92 Marks

In a simultanepous throw of a pair of dice, find the probability of getting:
A sum less than 7

Answer

Let E be the events that an even number on the first dice appear which means any number can be apper on second dice,
$\therefore\text{E} = \big\{(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(2, 1),(2, 2),(2, 3),\\\ \text{n(E)=15}(2,4),(3, 1),(3,2),(3,3),(4,1),(4,2),(5,1) \big\}$
$\text{p(E)}=\frac{15}{36}=\frac{5}{12}$

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Question 102 Marks

In a simultanepous throw of a pair of dice, find the probability of getting,A total greater than 8

Answer

Let E be the events that a total of 9 or 11 appear on the faces of dice.
$\therefore\ \text{E}= \big\{(3, 6),\ (4, 5),\ (4, 6),\ (5, 4),\ (5, 6),\ (6,3),\ (6, 4),\ (6, 5),\ (6, 6)\big\}$
$\therefore\text{n(E)}=10$
$\therefore\text{P(E)}=\frac{10}{36}=\frac{5}{18}$

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Question 112 Marks

$2$ boys and $2$ girls are in room $P$ and $1$ boy $3$ girls are in room $Q$. Write the sample space for the experiment in which a room is selected and then a person.

Answer

Let us denote two boys and two girls in room $P$ as $B_1, B_2$_ and $G_1, G_2$, respectively.
Let us denote one boy and three girls in room $Q$ as $B_3$ and $G_3, G_4, G_5$, respectively.
Accordingly, the required sample space is given by $S = {(P, B_1), (P, B_2), (P, G_1), (P, G_2) (Q, B_3), (Q, G_3), (Q, G_4), (Q, G_5)}.$

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Question 122 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:A diamond card

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$A diamond card
Let E be the events that A diamond cardappears,
$\therefore\ \text{n}\text{(E)} =\ ^{13}\text{C}_1=13$ $\therefore\ \text{P}\text{(E)}=\frac{13}{52}=\frac{1}{4}$

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Question 132 Marks

In a single throw of a die describe the following events:
F = Getting a number not less than 3.
Also, find A ∪ B, A ∩ B, B ∩ C, E ∩ F, D ∩ F and F.

Answer

When a dice is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.
Accordingly, we have:
F = {3, 4, 5, 6}
Here, F = {3, 4, 5, 6} and F = {1, 2, 3, 4, 5, 6}
$\therefore$ F = S - F = {1, 2,}

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Question 142 Marks

In a simultanepous throw of a pair of dice, find the probability of getting,A number greather than 4 on each die.

Answer

Let E be the events that a number greater than 4 appear on each dice.
$\therefore\ \text{E}=\big\{(5, 5),\ (5, 6),\ (6, 5),\ (6, 6)\big\}$
$\Rightarrow\text{n(E)}=4$
$\therefore\text{P(E)}=\frac{4}{36}=\frac{1}{9}$

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Question 152 Marks

In a hand at Whist, what is the probability that four kings are held by a specified player?

Answer

Since in one hand at whist a player has 13 cards
$\therefore\text{n}(\text{S})=^{52}\text{C}_{13}$
Let E be the event that a player has 4 kings
$\therefore\text{n}(\text{S})=^{4}\text{C}_{4}\times^{48}\text{C}_{9}$
$\therefore\text{p}(\text{E})=\frac{^{4}\text{C}_{4}\times^{48}\text{C}_{9}}{^{52}\text{C}_{13}}$
$=\frac{4\times^{48}\text{C}_{9}}{^{52}\text{C}_{13}}$
$=\frac{11}{4165}$

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Question 162 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:
A black king.

Answer

since a card is drawn form a pack of 52 cards.
Numbers of elementary events in the sample space
$\text{n}(\text{E})=\ ^{52}\text{C}_1=52$
A black king
$\therefore\text{n}(\text{E})=\ ^{2}\text{C}_1=2$
$\big[\therefore$There are two black kings spade and club kings$\big]$
$\therefore\text{P}(\text{E})=\frac{2}{52}=\frac{1}{26}$

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Question 172 Marks

Find the probability that in a random arrangement of the letters of the word 'UNIVERSITY', the two I's do not come together.

Answer

In the word 'UNIVERSITY' there are 10 letters. $\therefore\text{n}(\text{S})=10!$ Let E be the event that both the I's come together $\text{n}(\text{E})=2\times9!$ $\therefore\text{p}(\text{E})=\frac{2\times9!}{10!}=\frac{2}{10}=\frac{1}{5}$ $\therefore$ The probability that two I's do not come together is $\therefore\text{p}(\overline{\text{E}})=1-\text{p}(\text{E})=1-\frac{1}{5}=\frac{4}{5}$ $\therefore\text{p}(\overline{\text{E}})=\frac{4}{5}$

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Question 182 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:Black and a king.

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$Black and a king.
Let E be the events that a black and a king appear
$\therefore\text{n}(\text{E})=\ ^{2}\text{C}_1=2$ $\therefore\text{P}(\text{E})=\frac{2}{52}=\frac{1}{26}$

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Question 192 Marks

In a simultanepous throw of a pair of dice, find the probability of getting,Either 9 nor 11 as the sum of the numbers on the faces.

Answer

Let E be the events that either 9 nor 11 as the sum of the numbers on the faces of the dice.$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the events be the events that either 9 nor 11 as the sum of the numbers on the faces of the dice.
$\therefore\ \stackrel{{\sim}}{\hbox{E}}\ =\ \big\{(3, 6),\ (4, 5),\ (5, 4),\ (5, 6),\ (6, 3),\ (6, 5),\ (6, 4)\big\}$ $\therefore\ \text{n}\stackrel{{\sim}}{\hbox{(E)}}=6$ $ \text{p}\stackrel{{\sim}}{\hbox{(E)}}=\frac{6}{36}=\frac{1}{6}$ $\therefore\ \text{p(E)}=1-\text{p}\stackrel{{\sim}}{\hbox{(E)}}$ $=1-\frac{1}{6}=\frac{5}{6}$

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Question 202 Marks

An experiment consists of boy$-$girl composition of families with $2$ children.

What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
What is the sample space if we are interested in the number of boys in a family?

Answer

When the order of the birth of a girl or a boy is considered, the sample space is given by $\ce{S = {(G_1, G_2), (G_1, B_2), (B_1 ,G_2), (B_1, B_2)}}$
Since the maximum number of children in each family is two, a family can either have two boys or one boy or no boy.

Hence, the required sample space is given by $S = \{0, 1, 2\}.$

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Question 212 Marks

What is the probability that an ordinary year has 53 sundays?

Answer

Since in an ordinary year there are 52 weeks and one day.
so, we have to determine the probability of that one day being sunday.
$\text{S} = \big\{\text{M, T, W, T, H, F, S, S, U}\big\}$
$\therefore\text{p(E)}=\frac{1}{7} $

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Question 222 Marks

Two unbiased dice are thrown. find the probability that:

Neither a doublet nor a total of $8$ will appear
The sum of the numbers obtained on the two dice neither a multiple of $2$ nor a multiple of $3$.

Answer

Since two unbisased dice are thrown
$\therefore\text{n}\text{(S)}=6^2=36$
$(i)$ Let $E$ be the events that neither a doublet nor a total of $8$ will appear.
$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the event that a doublet or a total of $8$ will appear,
$\stackrel{{\sim}}{\hbox{E}}\ =\ \big\{(1,1),\ (2,2),\ (3,3)\ (4,4),\ (5,5),\ (6,6),\ (2,6)\ (3,5),\ (5,3),\ (6,2)\big\}$
$\therefore\ \text{n}\stackrel{{\sim}}{\hbox{(E)}}=10$
$\therefore\ \text{P}\stackrel{{\sim}}{\hbox{(E)}}=\frac{10}{36}$
$\text{P(E)}=1-\text{P}\stackrel{{\sim}}{\hbox{(E)}}$
$=1-\frac{10}{36}=\frac{26}{36}=\frac{13}{18}$
$(ii)$ Let $E$ be the events that the sum of the numbers obtained on the two dice is neither a multiple of $2$ nor a multiple of $3.$
$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the event that the sum of the numbers obtained on the two dice is either a multiple of $2$ or a multiple of $3,$ 5that is total shoud be $2, 3, 4, 6, 8, 9, 10, 12$
$\therefore\ \stackrel{{\sim}}{\hbox{E}}\ =\Big\{(1,1),\ (1,2),\ (2,1),\ (1,3),\ (2,2),\ (3,1),\ (1,5),\ (2,4),\ (3,3),\ (4,2),\ (5,1),\ (2,6),\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4,4),\ (5,3),\ (6,2),\ (3,6),\ (4,5),\ (5,4),\ (6,3),\ (4,6),\ (5,5),\ (6,4),\ (6,6)\Big\}$
$\therefore\ \text{n}\stackrel{{\sim}}{\hbox{(E)}}=24$
$\therefore\ \text{P}\stackrel{{\sim}}{\hbox{(E)}}=\frac{24}{36}$
$=\frac{4}{6}=\frac{2}{3}$
$\text{P(E)}=1-\text{P}\stackrel{{\sim}}{\hbox{(E)}}$
$=1-\frac{2}{3}=\frac{1}{3}$

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Question 232 Marks

What are the odds in favour of getting a spade if a card is drawn from a well $-$ shuffled deck of cards? what are the odds in favour of getting a king?

Answer

Let $E$ be the event of getting a spade from a

$\therefore\text{P(E)}=\frac{13}{52}=\frac{1}{4}$
$\Rightarrow\text{P}\bar{\text{(E)}}=\frac{3}{4}$
$\therefore$ Odds in favour of getting a spade from a pack of cards is
$\text{P(E)}:\text{P}\bar{\text{(E)}}=1:3$

Let $E$ be the event of getting aking from a pack of cards.

$\therefore\text{P(E)}=\frac{4}{52}=\frac{1}{13}$
$\Rightarrow\text{P}\bar{\text{(E)}}=\frac{12}{13}$
$\therefore$ Odds in favour of getting a king is,
$\text{P(E)}:\text{P}\bar{\text{(E)}}=1:12$

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Question 242 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:A black card

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$A black card
Let E be the events that A black card appears,
$\because$ There are 26 black cards (spade and club)
$\therefore\text{n}\text{(E)}=\ ^{26}\text{C}_1=26$ $\text{P}\text{(E)}=\frac{26}{52}=\frac{1}{2}$

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Question 252 Marks

Three coins are tossed Describe.Two events A and B which are not mutually exclusive.

Answer

When three coins are tossed, the sample space is given by,
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The two events that are not mutually exclusive a,
A: getting three heads
B: getting at least 2 heads
i.e. A = {HHH} and B = {HHH, HHT, HTH, THH}
This is because A ∩ B = {HHH} ≠ Φ

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Question 262 Marks

A die is thrown twice Each time the number appearing on it is recorded Describe the following events:
A = Both numbers are odd.

Answer

When a dice is thrown twice, we have the following possible outcomes:
A = both numbers are odd
= {(1, 1), (1, 3),(1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
Now, we have:
(A ∪ B) = {(1, 1), (1, 3), (1, 5) ,(3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5), (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}.

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Question 272 Marks

A card is picked up from a deck of 52 playing cards.
What is the event that the chosen card is a black faced card?

Answer

Let A be the event that the chosen card is a black faced card. Then,
A = {J♠, Q♠, K♠, J♣, Q♣, K♣}
Note: ♠ → Spade card, ♡ → Heart card, ♢→ Diamond card, ♣ → Club card

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Question 282 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:Not a diamond card

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$Not a diamond card
Let E be the events that not a diamond card appears,
$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the events that a diamond card appear
$\therefore\ \text{n}(\stackrel{{\sim}}{\hbox{E}})=\ ^{13}\text{C}_1=13$ $\therefore\ \text{P}\ (\stackrel{{\sim}}{\hbox{E}})=\frac{13}{52}=\frac{1}{4}$ $\therefore\ \text{P}\text{(E)}=1-\text{p}( \stackrel{{\sim}}{\hbox{E}})$ $=1-\frac{1}{4}=\frac{3}{4}$

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Question 292 Marks

A bag contains one white and one red ball. A ball is drawn from the bag. If the ball drawn is white it is replaced in the bag and again a ball is drawn. Otherwise, a die is tossed. Write the sample space for this experiment.

Answer

A bag contains one white ball (W) and one red ball (R).
When one ball is drawn, it will be either W or R.
The sample space of drawing one white ball with replacement and then again drawing a ball is {(W, W),(W, R)}.
Again, if red ball is drawn, a dice is rolled.
The sample space associated with this experiment is given by {(R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)}.
Hence, the sample space S for this experiment is S = {(W, W), (W, R), (R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)}.

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Question 302 Marks

A bag contains 2 red and white balls. three balls are drawn at random. find the probability that:
All the three balls are red.

Answer

Three balls are draw at random
$\therefore\ \text{n(S)}={^{13}\text{C}_3}$
(All the three balls are red)
$=\frac{{^8\text{C}_{3}}}{{^{13}{\text{C}}_3}}=\frac{\frac{8\times7\times6}{6}}{\frac{13\times12\times11}{6}}=\frac{8\times7\times6}{13\times12\times11}=\frac{28}{143}$

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Question 312 Marks

A coin is tossed repeatedly until a tail comes up for the first time. Write the sample space for this experiment.

Answer

In this experiment, a tail (T) may come up on the first throw, the second throw, the third throw and so on, until T is obtained.
This process continues indefinitely.
Hence, the sample space of this experiment is given by S = {T, HT, HHT, HHHT, HHHHT, HHHHHT, ...}

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Question 322 Marks

A bag contains 4 identical red balls and 3 identical black balls. The experiment consists of drawing one ball, then putting it into the bag and again drawing a ball. What are the possible outcomes of the experiment?

Answer

A bag contains four identical red balls (R) and three identical black balls (B).
The sample space S of drawing one ball with replacement and then again drawing a ball is given by,
S = {RR, RB, BR, BB}

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Question 332 Marks

One of the two events must happen. Given that the chance of one is two-third of the other, find the odds in favour of the other.

Answer

Let chance in favour of other be x
So $\text{x}+\frac{2}{3}\text{x}=1$
$\text{x}=\frac{3}{5}$
Odds in favour of other $=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}=3:2$

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Question 342 Marks

A bag contains 2 red and white balls. three balls are drawn at random. find the probability that:
One ball is red and two balls are white.

Answer

Three balls are draw at random
$\therefore\ \text{n(S)}={^{13}\text{C}_3}$
(One ball is red and two balls are white)
$=\frac{{^8\text{C}_{1}}\times^5\text{C}_2}{{^{13}{\text{C}}_3}}=\frac{8\times10}{\frac{13\times12\times11}{6}}=\frac{40}{143}$

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Question 352 Marks

A die is thrown twice Each time the number appearing on it is recorded Describe the following events:
C = sum of the numbers is less than
Also, find A ∪ B, A ∩ B, A ∪ C, A ∩ C Which pairs of events are mutually exclusive?

Answer

When a dice is thrown twice, we have the following possible outcomes:
C = sum of the numbers is less than 6
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
Now, we have
(A ∪ C) = {(1, 1), (1, 3),(1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5),
(1, 2), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(A ∩ C) = {(1, 1), (1, 3), (3, 1)}
Since (A ∩ B) = Φ and (A ∩ C) ≠ Φ, A and B are mutually exclusive, but A and C are not.

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Question 362 Marks

A coin is tossed twice. If the second draw results in a head, a die is rolled. Write the sample space for this experiment.

Answer

When a coin is tossed, the possible outcomes are head (H) and tail (T).
If a coin is tossed twice, the possible outcomes are given by,
S = {HH, HT, TH, TT}
Again, if the second draw results in a head, then a dice is rolled. The events associated with this experiment is given by,
A = {(HH, 1), (HH, 2), (HH, 3), (HH, 4), (HH, 5), (HH, 6), (TH, 1), (TH, 2), (TH, 3), (TH, 4), (TH, 5), (TH, 6)}
Hence, the sample space for this experiment is given by,
(S ∪ A) = {(HH, 1), (HH, 2), (HH, 3), (HH, 4), (HH, 5), (HH, 6), (TH, 1), (TH, 2), (TH, 3), (TH, 4), (TH, 5), (TH, 6), (HT), (TT)}.

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Question 372 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:A either a black card or a king.

Answer

since a card is drawn form a pack of 52 cards.
Numbers of elementary events in the sample space
$\text{n}(\text{E})=\ ^{52}\text{C}_1=52$
A either a black card or a king.
$\therefore\text{n}(\text{E})=\ ^{26}\text{C}_1+\ ^{4}\text{C}_1-\ ^2\text{C}_1=2$
$= 26+4-2 = 28$
$\big[\therefore$There are two black kings so we subtract in total$\big]$
$\therefore\text{P}(\text{E})=\frac{28}{52}=\frac{7}{13}$

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Question 382 Marks

The numbers $1, 2, 3$ and $4$ are written separately on four slips of paper. The slips are then put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement.
Describe the following events:
$A =$ The number on the first slip is larger than the one on the second slip.
$B =$ The number on the second slip is greater than $2.$
$C =$ The sum of the numbers on the two slips is $6$ or $7$.
$D =$ The number on the second slips is twice that on the first slip.
Which pair$(s)$ of events is $($are$)$ mutually exclusive?

Answer

Four slips marked as $1, 2, 3$ and $4$ are in a box. Two slips are drawn from it one after the other without replacement. The sample space S for the experiment is
$S = \{(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)\}$

$A =$ number on the first slip is larger than the one on the second slip

$= \{(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)\}$

$B =$ number on the second slip is greater than $2$

$= \{(1, 3), (2, 3), (1, 4), (2, 4), (3, 4), (4, 3)\}$

$C =$ sum of the numbers on the two slips is $6$ or $7$

$= \{(2, 4), (3, 4), (4, 2), (4, 3)\}$

$D =$ number on the second slip is two times the number on the first slip

$= \{(1, 2), (2, 4)\}$
Clearly, $(A ∩ D) = \Phi $
Therefore, $A$ and $D$ are mutually exclusive events.

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Question 392 Marks

A die is thrown repeatedly until a six comes up. What is the sample space for this experiment.

Answer

In this experiment, 6 may come up on the first throw, the second throw, the third throw and so on, until it is obtained.
Hence, the sample space of this experiment is given by,
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (2, 1, 6), (2, 2, 6), (2, 3, 6) …}

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Question 402 Marks

Fill in the blanks in the following table:

	$P(A)$	$P(B)$	$\text{P}({\text{A}}\cap{\text{B}})$	$\text{P}({\text{A}}\cup{\text{B}})$
$(i)$	$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	$.......$
$(ii)$	$0.35$	$....$	$0.25$	$0.6$
$(iii)$	$0.5$	$0.35$	$....$	$0.7$

Answer

Given,

$P(A) =\frac{1}{3},\ \text{P}(\text{A}\cap\text{B})=\frac{1}{15}$
$P(B) =\frac{1}{5},\ \text{P}(\text{A}\cup\text{B})=\ .......$
$\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$
$=\frac{5+3-1}{15}$
$=\frac{8-1}{15}=\frac{7}{15}$
$\because\text{P}(\text{A}\cup\text{B})=\frac{7}{15}$

Given,

$P(A) = 0.35, P(B) = ....$
$\text{P}(\text{A}\cap\text{B})=0.25,\ \text{P}(\text{A}\cup\text{B})=0.6$
$\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$0.6=0.35+\text{P}(\text{B})-0.25$
$0.6=0.10+\text{P}(\text{B})$
$\text{P}(\text{B})=0.6-0.1$
$\text{P}(\text{B})=0.5$

Given

$P(A) = 0.5, P(B) = 0.35$
$\text{P}(\text{A}\cap\text{B})=\ ...,\ \text{P}(\text{A}\cup\text{B})=0.7$
$\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$0.7=0.5+0.35-\text{P}(\text{A}\cap\text{B})$
$0.7=0.85-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{A}\cap\text{B})=0.85-0.7$
$\text{P}(\text{A}\cap\text{B})=0.15$

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Question 412 Marks

$A$ and $B$ are two events such that $P(A) = 0.54, P(B) = 0.69$ and $\text{P}(\text{A}\cap\text{B}) = 0.35.$ Find

$\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$\text{P}({\text{A}}\cap\overline{\text{B}})$
$\text{P}({\text{B}}\cap\overline{\text{A}})$

Answer

Given,
$P(A) = 0.54$
$P(B) = 0.69$
$\text{P}(\text{A}\cap\text{B})=0.35$

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

$=0.54+0.69-0.35$
$=1.23-0.35$
$\therefore\text{P}(\text{A}\cup\text{B})=0.88$

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$

$=1-0.88$
$=0.12$
$\therefore\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.12$

$\text{P}({\text{A}}\cap\overline{\text{B}})=\text{p}(\text{A})-\text{P}({\text{A}}\cap{\text{B}})$

$=0.54-0.35$
$=0.19$
$\therefore\text{P}({\text{A}}\cap\overline{\text{B}})=0.19$

$\text{P}({\text{B}}\cap\overline{\text{A}})=\text{P}(\text{B})-\text{P}({\text{A}}\cap{\text{B}})$

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

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Question 422 Marks

If odds against an event be 7 : 9, find the probability of non-occurrence of this event.

Answer

Since odd against an event is 7 : 9
$\therefore\text{n}\text{(S)}=7\text{K}+9\text{K}=16\text{K}$
Let E be the event that the event will occur
and, $\text{n}\text{(S)}=9\text{k}$
$\therefore\text{P}\text{(E)}=\frac{9}{16}$
$\therefore$ Probability of non-occurance of this event is
$\text{P(E)}=1-\text{P(E)}$
$=1-\frac{9}{1 6}$ $=\frac{7}{16}$

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Question 432 Marks

A box contains $1$ red and $3$ black balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.

Answer

The box contains one red ball and three black balls.
Let us denote the red ball as R and the three black balls as $B_1, B_2$ and $B_3$.
The sample space of this experiment is given by,
$S = {(R, B_1), (R, B_2), (R, B_3), (B_1, R), (B_1, B_2), (B_1, B_3), (B_2, B_1), (B_2, B_3), (B_2, R), (B_3, R), (B_3, B_1), (B_3, B_2)}.$

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Question 442 Marks

A box contains 1 white and 3 identical black balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.

Answer

It is given that the box contains one white ball and three identical black balls.
Let us denote the white ball with W and a black ball with B.
When two balls are drawn at random in succession without replacement, the sample space for this experiment will be given by S = {WB, BW, BB}.

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Question 452 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:Not an ace

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$Not a diamond card
Let E be the events that not an ace card does not appears,
$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the events that not an ace appears
$\therefore\ \text{n}(\stackrel{{\sim}}{\hbox{E}})=\ ^{4}\text{C}_1=4$ $\Rightarrow\ \text{P}\ (\stackrel{{\sim}}{\hbox{E}})=\frac{4}{52}=\frac{1}{13}$ $\therefore\ \text{P}\text{(E)}=\frac{1}{13}$ $\text{P}\text{(E)}=1-\frac{1}{13}=\frac{12}{13}$

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Question 462 Marks

In a single throw of three dice, find the probability of getting a total of 17 or 18.

Answer

$\therefore$ Three dice are thrown
$\therefore\text{n(s)}=6^3=216$
Let E be the event of getting total of if 17 or 18
$\therefore\ \text{E}=\big\{(6, 6, 5), \ (6, 5, 6),\ (5, 6, 6),\ (6, 6, 6),\ (6, 5),(6, 6)\big\}$
$\Rightarrow\text{n(E)}=4$
$\therefore\text{P(E)}=\frac{\text{n(E)}}{\text{n(s)}}$
$=\frac{4}{216}=\frac{1}{54}$
$\therefore\text{p(E)}=\frac{1}{54}$

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Question 472 Marks

A committee of two person is selected from two men and two women. what is the probebility that the committee will have,

No man?
One Man?
Two men?

Answer

We have, Two men and two women
Now, a committee of two person is selected
$\therefore\text{n}\text{(S)}=\ ^{4}\text{C}_2=\frac{4\times3}{2}=6$
(i) Let $E$ be the event that no man is to be in committee
$\therefore\text{n}\text{(E)}=\ ^{2}\text{C}_2=1$
[only woman will be the committe]
$\therefore\text{P}\text{(E)}=\frac{1}{6}$
(ii) Let $E$ be the event that one man is to be in committee
$\therefore\text{E}=\text{(m,10)}$
$\therefore\text{n}\text{(E)}=\ ^{2}\text{C}_1\times\ ^{2}\text{C}_1$
$=2\times2=4$
$\therefore\text{P}\text{(E)}=\frac{4}{6}=\frac{2}{3}$
(iii) Let $E$ be the event that two man in the committee
$\therefore\text{n}\text{(E)}=\ ^{2}\text{C}_2-1$
$\therefore\text{P}\text{(E)}=\frac{1}{6}$

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Question 482 Marks

Two unbisased dice are thrown, find the probability that the total of the numbers on the dice is greater than 10.

Answer

$\because$ Two dice are thrown
$\because\text{n}\text{(S)} = 6^2 = 36$
Let E be the event of getting that of the numbers on the dice is greater than 10
$\therefore\text{E} = \big\{\text{(5, 6), (6, 5), (6, 6)}\big\}$
$\therefore\text{n(E)=3}$
$\therefore\text{p(E)}=\frac{3}{36}=\frac{1}{12}$
$\therefore\text{p(E)}=\frac{1}{36}$

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Question 492 Marks

Five cards are drawn from a well-shuffled pack of 52 cards. Find the probability that all the five cards are hearts.

Answer

Five cards are drawn from a well schuffled pack of cards
$\therefore\text{n}(\text{S})=^{52}\text{C}_5$
Let E be the event that all the five cards are hearts
$\therefore\text{n}(\text{E})=^{13}\text{C}_5$
$\therefore\text{p}(\text{E})=\frac{^{13}\text{C}_5}{^{52}\text{C}_5}$
$=\frac{13\times12\times11\times10\times9}{52\times51\times50\times49\times48}$
$=\frac{33}{66640}$

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Question 502 Marks

Three coins are tossed Describe,
Two events A and B which are mutually exclusive but not exhaustive.

Answer

When three coins are tossed, the sample space is given by,
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The two events which are mutually exclusive but not exhaustive are as follows:
A: getting exactly one head
B: getting exactly one tail
i.e. A = {HTT, THT, TTH} and B = {HHT, HTH, THH}
It is because, A ∩ B = Φ, but A ∪ B ≠ S

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