3 Marks Question

Question 13 Marks

A group consists of $12$ persons, of which $3$ are extremely patient, other $6$ are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each persion is equally likely to be selected, find the probability of selecting a person who is:

Extermely patient.
Extremely kind or honest. Which of the above values you prefer more.

Answer

The total number of persons in the group is $12$ out of which $3$ are extermely patient patient, other $6$ are extremely hoest and $3$ are extermely kind.

$P($selectcting an extermely patient person$)$

$=\frac{3}{12}$
$=\frac{1}{4}$

There are $9$ persons who are either extermely kind or extermely honest.

$P($selecting an extremely kind or honest person$)$
$=\frac{9}{12}$
$=\frac{3}{4}$

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Question 23 Marks

A letter is chosen at random from the letters of the word $\text{'ASSOCIATION'}.$ Find the probability that the chosen letter is a:

Vowel.
Consonant.
An $S.$

Answer

There are $11$ letters in the word $\text{ASSOCIATION.}$
Total number of outcomes $= 11$
The vowels are $A, 0$ and $1$ appear twice each
So, there are $6$ letter which are vowels in the given word.
So, there are $11 - 6 = 5$ consonants

$P($getting a vowel$) =\frac{6}{11}$
$P($getting a consonant$) =\frac{5}{11}$
There are two $S$ letters.

$P($getting an $S) =\frac{2}{11}$

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Question 33 Marks

A box contains $90$ discs which are numbered from $1$ to $90$. If one disc is drawn at random from the box, find the probability that it bears:

A two-digit number.
A perfect square number.
A number divisible by $5.$

Answer

Total number of discs $= 90$

The two-di git numbers would be $10, 11, 12, ..., 90.$
So, there are $81$ two-digit numbers.
$P($getting a two-digit number$)$
$=\frac{81}{90}$
$=\frac{9}{10}$

The perfect squares form $1$ to $90$ are $1, 4, 9, 16, 25, 36, 49, 64$ and $81.$

So, there are $9$ perfect squares.
$P($getting a perfect square$)$
$=\frac{9}{90}$
$=\frac{1}{10}$

The number divisible by $5$ are $5, 10, 15, 25, 30, 35, 50, 55, 60, 65, 70, 75, 80, 85,$ and $90.$

$P($getting a number divisible by $5)$
$=\frac{18}{90}$
$=\frac{1}{5}$

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Question 43 Marks

A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball find the number of black balls in the bag.

Answer

White ball $= 15$
black ball $= x$
Total outcome $= 15 + x$
$P_1$ (of white ball) $=\frac{15}{15+\text{x}}$
$P_2$ (of black ball) $=\frac{\text{x}}{15+\text{x}}$
$P_2= 3P$
$\Rightarrow\frac{\text{x}}{15+\text{x}}=3\times\frac{15}{15+\text{x}}$
$\Rightarrow\frac{\text{x}}{15+\text{x}}=\frac{45}{\text{x}+15}$
$\Rightarrow\text{x}=45$

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Question 53 Marks

A game of chance consists of spinning an arrow, which comes to rest pointing at one of the number $1, 2, 3, 4, 5, 6, 7, 8$ and these are equally likely outcomes. Find the probability that the arrow will point at any factor of $8.$

Answer

Total number of outcomes $= 8$
The factors of $8$ are $1, 2, 4$ and $8.$
So, there are $4$ possible outcomes.
$P($getting a factor of $8)$
$=\frac{4}{8}$
$=\frac{1}{2}$
Note: The answer given in the text in incorrect.
The correct answer is as shown above.

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Question 63 Marks

Cards marked with numbers $1, 3, 5, ..., 101$ are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn card is:

Less than $19.$
A prime number less than $20.$

Answer

To find the number of cards in the bag, we use the general formula for an $AP$ since the number on the cards are in $AP.$
Here, first term, $a = 1$, common difference $= d = 2$
Let n be the number of cards in the bag.
$a_n= a + (n - 1)d$
$⇒ 101 = 1 + (n - 1)(2)$
$⇒ 101 = 2n - 2$
$⇒ 102 = 2n$
$⇒ n = 51$
So, there $51$ cards in the bag.

The number less than $19$ are $1, 3, 5, 7, 9, 11, 13, 15$ and $17.$

So, there are $9$ possible outcomes.
$P($getting a number less than $19)$
$=\frac{9}{51}$
$=\frac{3}{17}$

The primes less than $20$ are $3, 5, 7, 77, 13, 17, 19.$

So, there are $7$ possible oucomes.
$P($getting a prime less than $20)$
$=\frac{7}{51}$

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Question 73 Marks

Five cards the ten, jack, queen, king and ace of diamonds are well shuffled with their faces downwards. One card is then picked up at random.

What is the probability that the drawn card is the queen?
If the queen is drawn and put aside and a second card is drawn, find the probability that the second card is: $(i)$ An ace. $(ii)$ A queen.

Answer

Given that there are $5$ cards.
Total number of outcomes $= 5$

There is $1$ queen.

$P($getting a queen$)$
$=\frac{1}{5}$

Given that a queen is drawn and put asi de.

So, there will be $4$ cards from where the ace will be chosen

$P($getting an ace$)$

$=\frac{1}{4}$

Since the queen is kept aside, there is no queen left.

$P($getting a queen$) = 0$

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Question 83 Marks

A bag contains lemon-flavoured condies only. Hema takes out one condy without looking into the bag. What is the probability that she takes out

An orange-flavoured candy$?$
A lemon-flavoured candy$?$

Answer

$P($that she takes out an orange-flavoured candy$) = 0$

$($Since there are no orange-flavoured candies in the bag$)$

$P($that she takes out a lemon-flavoured candy$) = 1$

$($Since there are only lemon-flavoured candies in the bag$)$

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Question 93 Marks

A card is drawn at random from a well-shuffled pack of $52$ cards. Find the probability that the drawn card is neither a king nor a queen.

Answer

There are $26$ red cards containing a $2$ queensand $2$ more black queens are there in a pack of cards
$\therefore P($getting a red card or a queen$)$
$=\frac{28}{52}$
$=\frac{7}{13}$
$\therefore P($getting neither a red card nor a queen$)$
$=\Big(1-\frac{7}{13}\Big)$
$=\frac{6}{13}$

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Question 103 Marks

A piggy bank contains hundred $50-p$ coins, seventy Rs. $1$ coin, fifty Rs. $2$ coins and thirty Rs. $5$ coins. If it is equally likely that one of the coins will fall out when the bank is turned upsic down, what is the probability that the coin.

Will be a $Rs. 1$ coin.
Will not be a $Rs. 5$ coin.
Will be $50-p$ or a $Rs. 2$ coin.

Answer

The total number of coins in the piggy bank $= 100 + 70 + 50 + 30 = 250$

There are sevent $Rs. 1$ coins.

So, there are $70$ possible outcomes.
$p($getting a $Rs. 1$ coin$)$
$=\frac{70}{250}$
$=\frac{7}{25}$

There are thirty $Rs. 5$ coins.

So, the number of coins that is not a $Rs. 5$ coins are $250 - 30 = 220$
So, there are $220$ possible outcomes.
$P($getting a coin that is not a $Rs. 5$ coin$)$
$=\frac{220}{250}$
$=\frac{22}{25}$

The number of $50-p$ coins are $100$ and the number of $Rs. 2$ coin s are $50.$

So, there are $100 + 50 = 150$ possible outcomes.
$P($getting a coin that will be $50-p$ or a $Rs. 2$ coin$)$
$=\frac{150}{250}$
$=\frac{3}{5}$

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Question 113 Marks

A card is drawn at random from a well-shuffled pack of $52$ cards. Find the probability of getting

A red king.
A queen or a jack.

Answer

There are $52$ cards in the pack.

There are $2$ red kings

$P($getting a red king$)$
$=\frac{2}{52}$
$=\frac{1}{26}$

There are $4$ queens and $4$ jacks

So, there are 8 possible outcomes.
$P($getting a queen or jack$)$
$=\frac{8}{52}$
$=\frac{2}{13}$

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Question 123 Marks

A jar contains $24$ marbles. Some of these are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is $\frac{2}{3}.$ Find the number of blue marbles in the jar.

Answer

Total number of marbles in the jar $= 24$
i.e. $n(S) = 24$
Let number of blue balls in the jar be $x.$
Then, the number of green balls in the jar $= 24 - x$
Let $A$ be the favourable outcomes of getting blue balls.
Then, $n(A) = x$
Therefore, $P(A)$
$=\frac{\text{n(A)}}{\text{n(S)}}$
$=\frac{\text{x}}{24}$
Let $B$ be favourable outcomes of getting green balls.
Then, $n(B) = 24 - x$
Therefore, $P(B)$
$=\frac{\text{n(B)}}{\text{n(S)}}$
$=\frac{24-\text{x}}{24}$
According to the given condition: $P(B) =\frac{2}{3}$
$\Rightarrow\frac{24-\text{x}}{24}=\frac{2}{3}$
$\Rightarrow3(24-\text{x})=2\times24$
$\Rightarrow72-3\text{x}=48$
$\Rightarrow-3\text{x}=-24$
$\Rightarrow\text{x}=8$
Hence, the number of blue balls in the jar $= 8.$

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Question 133 Marks

A card is drawn from a well-shuffled pack of $52$ cards. Find the probability of getting:

A red face card.
A black king.

Answer

There are $52$ cards in the pack.

There are $6$ red face cards.

$P($getting a red face card$)$
$=\frac{6}{52}$
$=\frac{3}{26}$

There are $2$ black kings

$P($getting a black king$)$
$=\frac{2}{52}$
$=\frac{1}{26}$

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