M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip

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M.C.Q (1 Marks)

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37 questions · auto-graded multiple-choice test.

MCQ 11 Mark

Two coins are tossed simultaneously. What is the probability of getting at most one head?

A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$\frac{2}{3}$
✓
$\frac{3}{4}$

Answer

Correct option: D.

$\frac{3}{4}$

When two coins are tossed the simultanecusly the
outcomes are:
$\text{\{HH, HT, TH, TT\}}$
So, there are 4 outcomes.
Getting atmost one head means the possible outcomes are:
$\text{\{HT, TH, TT\}}$
So, there are $3$ possible outcomes.
$P($getting atmost one head$)$
$=\frac{3}{4}$

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MCQ 21 Mark

Cards, each marked with one of the numbers $6, 7, 8, ..., 15$ are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with number less than $10?$

A
$\frac{3}{5}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
✓
$\frac{2}{5}$

Answer

Correct option: D.

$\frac{2}{5}$

The total number of tickets $= 10$
The numbers less than $10$ are $6, 7, 8$ and $9.$
So, there are $4$ numbers.
$P($getting a number less than $10)$
$=\frac{4}{10}$
$=\frac{2}{5}$

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MCQ 31 Mark

A die is thrown once. The probability of getting an odd number greater than $3$ is:

A
$\frac{1}{3}$
✓
$\frac{1}{6}$
C
$\frac{1}{2}$
D
$0$

Answer

Correct option: B.

$\frac{1}{6}$

The number on a die are $1, 2, 3, 4, 5$ and $6.$
So, there are $6$ numbers in total.
The odd number on a die greater than $3$ is $5.$
So, there is only $1$ number.
$P($getting an odd number greater than $3)$
$=\frac{1}{6}$

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MCQ 41 Mark

The probability of throwing a number greater than $2$ with a fair die is:

A
$\frac{2}{5}$
B
$\frac{5}{6}$
C
$\frac{1}{3}$
✓
$\frac{2}{3}$

Answer

Correct option: D.

$\frac{2}{3}$

The numbers on a fair die are $1, 2, 3, 4, 5$ and $6.$
So, there are $6$ numbers in total.
The number greater than $2$ are $3, 4, 5$ and $6.$
So, there are $4$ numbers.
$P($getting a number greater than $2)$
$=\frac{4}{6}$
$=\frac{2}{3}$

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MCQ 51 Mark

One card is drawn at random from a well-shuffled deck of $52$ cards. What is th probability of getting a face card?

A
$\frac{1}{26}$
B
$\frac{3}{26}$
✓
$\frac{3}{13}$
D
$\frac{4}{13}$

Answer

Correct option: C.

$\frac{3}{13}$

The total number of cards $= 52$
The number of queens $= 12$
$P($getting a face card$)$
$=\frac{12}{52}$
$=\frac{3}{13}$

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MCQ 61 Mark

A card is drawn at random from a well$-$shuffled deck of $52$ cards. What is the probability of getting a black king?

A
$\frac{1}{13}$
✓
$\frac{1}{26}$
C
$\frac{2}{39}$
D
$\text{None of these}$

Answer

Correct option: B.

$\frac{1}{26}$

The total number of cards $= 52$
The number of black kings $= 2$
$P($getting a black king$)$
$=\frac{2}{52}$
$=\frac{1}{26}$

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MCQ 71 Mark

Which of the following cennot be the probability of an event?

✓
$1.5$
B
$\frac{3}{5}$
C
$25\%$
D
$0.3$

Answer

Correct option: A.

$1.5$

We know that, the probability of an event $E$ will always lie betweent $0$ and $1.$
Since $1.5 > 1,$ it cannot be the probability of an event.

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MCQ 81 Mark

If an event cannot occur then its probability is:

A
$1$
B
$\frac{1}{2}$
C
$\frac{3}{4}$
✓
$0$

Answer

Correct option: D.

$0$

An event that cannot ocour is called an impossibel event.
the probability of an impossible event is $0.$

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MCQ 91 Mark

A die is thrown once. The probability of getting a prime number is:

A
$\frac{2}{3}$
B
$\frac{1}{3}$
✓
$\frac{1}{2}$
D
$\frac{1}{6}$

Answer

Correct option: C.

$\frac{1}{2}$

The number on a die are $1, 2, 3, 4, 5$ and $6.$
So, there are $6$ numbers in total.
The prime numbers on the die are $2, 3,$ and $5.$
So, there are $3$ numbers.
$P($getting a prime number on the die$)$
$=\frac{3}{6}$
$=\frac{1}{2}$

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MCQ 101 Mark

The probability of getting $2$ heads, when two coins are tossed, is:

A
$1$
B
$\frac{3}{4}$
C
$\frac{2}{2}$
✓
$\frac{1}{4}$

Answer

Correct option: D.

$\frac{1}{4}$

When two coins are tossed the outcomes are:
$\text{\{HH, HT, TH, TT\}}$
So, there are $4$ numbers in total
$P($getting one head$)$
$=\frac{1}{4}$

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MCQ 111 Mark

One ticket is drawn at random from a bag containing tickets numbered $1$ to $40$. The probability that the selected ticket has a number, which is a multiple of $7,$ is:

A
$\frac{1}{7}$
✓
$\frac{1}{8}$
C
$\frac{1}{5}$
D
$\frac{7}{40}$

Answer

Correct option: B.

$\frac{1}{8}$

The total number of tickets $= 40$
The multiples of $7$ between $1$ and $40$ are $7, 14, 21, 28$ and $35.$
So. there are $5$ numbers.
$P($getting a multiple of $7)$
$=\frac{5}{40}$
$=\frac{1}{8}$

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MCQ 121 Mark

In a lottery, there are $6$ prizes and $24$ blanks. What is the probability of not getting a prize?

A
$\frac{3}{4}$
B
$\frac{3}{5}$
✓
$\frac{4}{5}$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{4}{5}$

The number of prizes $= 6$
The number of blanks $= 24$
So, the total number of tickets $= 6 + 24 = 30$
$P($not getting a prize$)$
$=\frac{24}{30}$
$=\frac{4}{5}$

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MCQ 131 Mark

The probability that a number selected at random from the numbers $1, 2, 3, ..., 15$ is a multiple of $4$ is:

A
$\frac{4}{15}$
B
$\frac{2}{15}$
✓
$\frac{1}{5}$
D
$\frac{1}{3}$

Answer

Correct option: C.

$\frac{1}{5}$

The selected numbers would be $4, 8$, and $12.$
So, there are $3$ number.
$P($number of multiples of $4)$
$=\frac{\text{Number of multipes of 4}}{\text{Total}}$
$=\frac{3}{15}$
$=\frac{1}{5}$

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MCQ 141 Mark

A die is thrown once. The probability of getting an even number is:

✓
$\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{1}{6}$
D
$\frac{5}{6}$

Answer

Correct option: A.

$\frac{1}{2}$

The numbers on a die are $1, 2, 3, 4, 5$ and $6.$
So, there are 6 numbers in total.
The even number on the die are $2, 4$ and $6.$
So, there are $3$ even number.
$P($getting an even number$)$
$=\frac{3}{6}$
$=\frac{1}{2}$

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MCQ 151 Mark

A bag contains $4$ red and $6$ black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball?

A
$\frac{2}{5}$
✓
$\frac{3}{5}$
C
$\frac{1}{10}$
D
$\text{None of these}$

Answer

Correct option: B.

$\frac{3}{5}$

The bag contains $4$ red and $6$ black balls.
So, the total number of balls $= 4 + 6 = 10$
The number of black balls $= 6$
$P($getting a black ball$)$
$=\frac{6}{10}$
$=\frac{3}{5}$

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MCQ 161 Mark

What is the probability of a sure event?

A
$0$
B
$\frac{1}{2}$
✓
$1$
D
Less than $1$

Answer

Correct option: C.

$1$

The probability of a sure event is always $1.$

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MCQ 171 Mark

What is the probability of an impossible event?

A
$\frac{1}{2}$
✓
$0$
C
$1$
D
More than $1$

Answer

Correct option: B.

$0$

The probability of an impossible event is always $0.$

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MCQ 181 Mark

A bag contains $3$ white, $4$ red and $5$ black balls. One ball is drawn at random. What is the probability that the ball drawn is neither black nor white?

A
$\frac{1}{4}$
B
$\frac{1}{2}$
✓
$\frac{1}{3}$
D
$\frac{3}{4}$

Answer

Correct option: C.

$\frac{1}{3}$

The bag contains $3$ white, $4$ red and $5$ black and balls.
So, the total number of balls $= 3 + 4 + 5 = 12$
For the ball that is drawn to be neither black not white, it should be red.
The number of red balls $= 4$
$P($getting a red ball$)$
$=\frac{4}{12}$
$=\frac{1}{3}$

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MCQ 191 Mark

Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

A
$\frac{1}{2}$
B
$\frac{1}{4}$
✓
$\frac{3}{8}$
D
$\frac{3}{4}$

Answer

Correct option: C.

$\frac{3}{8}$

When three coins are tossed the simultaneously the
outcomes are:
$\text{\{HHH, HHT, HTH, THT, HTT, TTH and TTT\}}$
So, there are $8$ possible outcones.
$P($getting exactly two heads$)$
$=\frac{3}{8}$

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MCQ 201 Mark

Two dice are thrown together. The probability of getting a doulet is:

A
$\frac{1}{3}$
✓
$\frac{1}{6}$
C
$\frac{1}{4}$
D
$\frac{2}{3}$

Answer

Correct option: B.

$\frac{1}{6}$

The number on each die are $1, 2, 3, 4, 5$ and $6.$
So, the total possibilities are:

$(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$

$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$

So, there are $36$ number in toral.
There are 6 possibilities when we obtain a doublet,
$(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).$
$P($getting a doublet$)$
$=\frac{6}{36}$
$=\frac{1}{6}$

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MCQ 211 Mark

If $P(E)$ denotes the probability of an $E$ then:

A
$\text{P(E)}<0$
B
$\text{P(E)}<1$
✓
$0\leq\text{P(E)}\leq1$
D
$-1\leq\text{P(E)}\leq1$

Answer

Correct option: C.

$0\leq\text{P(E)}\leq1$

We know that, the probability of an event $E$ will always lie between $0$ and $1,$
Where $0$ is the probability of an impossible event and $1$ is the probability of a sure event.

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MCQ 221 Mark

Two dice are thrown together. The probabililty of getting the same number on both dice is:

A
$\frac{1}{2}$
B
$\frac{1}{3}$
✓
$\frac{1}{6}$
D
$\frac{1}{12}$

Answer

Correct option: C.

$\frac{1}{6}$

The number on each die are $1, 2, 3, 4, 5$ and $6.$
So, the total possibilities are:

$(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$

$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$

So, there are $36$ number in toral.
There are $6$ possibilities when the two die
have the same number $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).$
$P($getting the same number on both the die$)$
$=\frac{6}{36}$
$=\frac{1}{6}$

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MCQ 231 Mark

If the probability of winning a game $0.4$ then the probability of losing it, is:

A
$0.96$
B
$\frac{1}{0.4}$
✓
$0.6$
D
None of these

Answer

Correct option: C.

$0.6$

We know that, if $E$ is an event, then $P(E) + P(E') = 1.$
Let $E$ be the event where the game is won.
So, $0.4 + P(E') = 1$
$\Rightarrow P(E') = 1 - 0.4$
$\Rightarrow P(E') = 0.6$
So, the probability of losing the game is $0.6.$

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MCQ 241 Mark

There are $20$ tickets numbered as $1, 2, 3, ..., 20$ respectively. One ticket is drawn at random. what is the probability that the number on the ticket drawn is a multiple of $5?$

A
$\frac{1}{4}$
✓
$\frac{1}{5}$
C
$\frac{2}{5}$
D
$\frac{3}{10}$

Answer

Correct option: B.

$\frac{1}{5}$

The total number of tickets $= 20$
The multiples of $5$ between $1$ and $20$ are $5, 10, 15$ and $20.$
So, there are $4$ numbers.
$P($getting a multiple of $5)$
$=\frac{4}{20}$
$=\frac{1}{5}$

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MCQ 251 Mark

A box contains $90$ discs, numbered from $1$ to $90$. If one disc is drawn at random from the bex, the probability that it bears prime number less than $23$ is:

A
$\frac{7}{90}$
B
$\frac{1}{9}$
✓
$\frac{4}{45}$
D
$\frac{8}{89}$

Answer

Correct option: C.

$\frac{4}{45}$

The total number of discs $= 90$
The primes less than $23$ are $2, 3, 5, 7, 11, 13, 17, 19,$
So. there are $8$ numbers.
$P($getting a prime number less than $23)$
$=\frac{8}{90}$
$=\frac{4}{45}$
Note: in the text, the option $(c)$ is incorrect.
it should be $\frac{4}{45}$ to go with the question asked.

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MCQ 261 Mark

If the probability of occurrence of an event is $p$ then the probability of non$-$happening of this event is:

A
$(p - 1)$
✓
$(1 - p)$
C
$p$
D
$\Big(1-\frac{1}{\text{p}}\Big)$

Answer

Correct option: B.

$(1 - p)$

Let $E$ be the event.
So, the probability of the event happening will be $P(E).$
Thus, the probability of the event not happening will be $P(E').$
Given that, $P(E) = p$
We know that, $P(E) + p(E') = 1$
$\Rightarrow p + P(E') = 1$
$\Rightarrow P(E') = 1 - p$

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MCQ 271 Mark

One card is drawn at random from a well$-$shuffled deck of $52$ cards. What is the probability of getting a black face card?

A
$\frac{1}{26}$
✓
$\frac{3}{26}$
C
$\frac{3}{13}$
D
$\frac{3}{14}$

Answer

Correct option: B.

$\frac{3}{26}$

The total number of cards $= 52$
The number of black face cards $= 6$
$P($getting a black face card$)$
$=\frac{6}{52}$
$=\frac{3}{26}$

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MCQ 281 Mark

Which of the following cannot be the probability of an event?

A
$\frac{1}{3}$
B
$0.3$
C
$33\%$
✓
$\frac{7}{6}$

Answer

Correct option: D.

$\frac{7}{6}$

We know that, the probability of an event $E$ will always lie between $0$ and $1.$
Since $\frac{7}{6}>1,$ it cannot be the probability of an event.

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MCQ 291 Mark

From a well$-$shuffled deck of $52$ cards, one card is drawn at random. What is the probability of getting a queen?

✓
$\frac{1}{13}$
B
$\frac{1}{26}$
C
$\frac{4}{39}$
D
$\text{None of these}$

Answer

Correct option: A.

$\frac{1}{13}$

The total number of cards $= 52$
The number of queens $= 4$
$P($getting a queen$)$
$=\frac{4}{52}$
$=\frac{1}{13}$

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MCQ 301 Mark

There are $25$ tickets numbered as $1, 2, 3, 4, ..., 25$ respectively. One ticket is drawn at random. what is the probability that the number on the ticket is a multiple of $3$ or $5?$

A
$\frac{2}{5}$
B
$\frac{11}{25}$
✓
$\frac{12}{25}$
D
$\frac{13}{25}$

Answer

Correct option: C.

$\frac{12}{25}$

The total number of tickets $= 25$
The multiples of $3$ are $3, 6, 9, 12, 15, 18, 21, 24.$
The multiples of $5$ are $5, 10, 15, 20$ and $25.$
Since $15$ is a multiple of $3$ as $5$, it is to be caculated only once.
So, there are $12$ numbers
$P($getting a multiple of $3$ or $5)$
$=\frac{12}{25}$

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MCQ 311 Mark

A number is selected at random from the nubers $1$ to $30$. What is the probability that the selected number is a prime number?

A
$\frac{2}{3}$
B
$\frac{1}{6}$
✓
$\frac{1}{3}$
D
$\frac{11}{30}$

Answer

Correct option: C.

$\frac{1}{3}$

The prime numbers from $1$ to $30$ are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29.$
So, there are $10$ prime numbers between $1$ and $30.$
$P($getting a prime number$)$
$=\frac{\text{Number of primes between 1 and 30}}{\text{Total}}$
$=\frac{10}{30}$
$=\frac{1}{3}$

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MCQ 321 Mark

A bag contains $8$ red, $2$ black and $5$ white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?

A
$\frac{8}{15}$
B
$\frac{2}{15}$
✓
$\frac{13}{15}$
D
$\frac{1}{3}$

Answer

Correct option: C.

$\frac{13}{15}$

The bag contains $8$ red, $2$ black and $5$ white balls.
So, the total number of balls $= 8 + 2 + 5 = 15$
Since the ball should not be black, it can be red or white.
The number of red and white balls $= 13$
$P($getting a red and white ball$)$
$=\frac{13}{15}$

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MCQ 331 Mark

Cards bearing numbers $2, 3, 4, ..., 11$ are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is:

✓
$\frac{1}{2}$
B
$\frac{2}{5}$
C
$\frac{3}{10}$
D
$\frac{5}{9}$

Answer

Correct option: A.

$\frac{1}{2}$

Number on the cards are $2, 3, 4, ..., 11.$
Sample space associated with the experiment, $S = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$
$\therefore\ $Total number of outcomes $= 10$
Let a be the event of drawing a card with a prime number.
The cards with prime number are $2, 3, 5, 7$ and $11.$
Number of outcomes in favour of event $A = 5$
$\therefore$ Requried probability $P(A)$
$=\frac{\text{Number of outcomes in favour of A }}{\text{Total number of outcomes i}}$
$=\frac{5}{10}$
$=\frac{1}{2}$

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MCQ 341 Mark

One card is drawn at random from a well$-$shuffled deck of $52$ cards. What is the probability of getting a $6?$

A
$\frac{3}{26}$
B
$\frac{1}{52}$
✓
$\frac{1}{13}$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{1}{13}$

The total number of cards $= 52$
The number of $6$ in the deck of cards $= 4$
$P($getting a $6)$
$=\frac{4}{52}$
$=\frac{1}{13}$

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MCQ 351 Mark

In a lottery, there are $8$ prizes and $16$ blanks. What is the probability of getting a prize?

A
$\frac{1}{2}$
✓
$\frac{1}{3}$
C
$\frac{2}{3}$
D
$\text{None of these}$

Answer

Correct option: B.

$\frac{1}{3}$

The number of prizes $= 8$
The number of blanks $= 16$
So, the total number of tickets $= 8 + 16 = 24$
$P($getting a prize$)$
$=\frac{8}{24}$
$=\frac{1}{3}$

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MCQ 361 Mark

A box contains $3$ blue, $2$ white and $4$ red marbles. If a marble is drawn at random from the box, what is the probability that it will not be a white marble?

A
$\frac{1}{3}$
B
$\frac{4}{9}$
✓
$\frac{7}{9}$
D
$\frac{2}{9}$

Answer

Correct option: C.

$\frac{7}{9}$

The bag contains $3$ blue, $2$ white and $4$ red marbles.
So, the total number of marbles $= 3 + 2 + 4 = 9$
Since the marbles cannot be white, it can blue or red.
The number of blue or red marbles $= 3 + 4 = 7$
$P($getting a blue or red marble$)$
$=\frac{7}{9}$

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MCQ 371 Mark

A box contains cards numbered $6$ to $50.$ A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square is:

A
$\frac{1}{45}$
B
$\frac{2}{15}$
C
$\frac{4}{45}$
✓
$\frac{1}{9}$

Answer

Correct option: D.

$\frac{1}{9}$

The numbers on the card have to be perfect sqaures.
So, the numbers would be $9, 16, 25, 36, 49.$
So, there are $5$ numbers
Total number of cards $= (50 - 6) + 1$
$= 44 + 1$
$= 45$
$P($getting a perfect square$)$
$=\frac{\text{Number of perfect squares}}{\text{Total}}$
$=\frac{5}{45}$
$=\frac{1}{9}$

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