M.C.Q (1 Marks)

MCQ 1511 Mark

Two dice are thrown simultaneously. The probability of getting a pair of aces is

✓
$\frac{1}{36}$
B
$\frac{1}{3}$
C
$\frac{1}{6}$
D
None of these.

Answer

Correct option: A.

$\frac{1}{36}$

Required probability = Probability of ace in first throw + Probability of ace in second throw

$=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

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MCQ 1521 Mark

The probabilities of a student getting I, II and III division in an examination are $\frac{1}{10},\frac{3}{5}$ and $\frac{1}{4}$ respectively. The probability that the student fails in the examination is.

A
$\frac{197}{200}$
✓
$\frac{27}{100}$
C
$\frac{83}{100}$
D
None of these.

Answer

Correct option: B.

$\frac{27}{100}$

$\text{P(A)}=\frac{1}{10},\text{P(B)}=\frac{3}{5},\text{P(C)}=\frac{1}{4}$
Required probability $=\text{P}(\overline{\text{A}}\cap\overline{\text{B}}\cap\overline{\text{C}})$

Required probability $=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})\text{ P}(\overline{\text{C}})$

Required probability $=(1-\text{P(A)})(1-\text{P(B)})(1-\text{P(C)})$

Required probability $=\Big(1-\frac{1}{10}\Big)\Big(1-\frac{3}{5}\Big)\Big(1-\frac{1}{4}\Big)$

Required probability $=\frac{27}{100}$

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MCQ 1531 Mark

A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is:

A
$\frac{\text{ }^{20}\text{C}_{10}\times5^6}{6^{20}}$
B
$\frac{120\times5^7}{6^{10}}$
✓
$\frac{84\times5^6}{6^{10}}$
D
None of these

Answer

Correct option: C.

$\frac{84\times5^6}{6^{10}}$

A fair die is thrown then probebility of getting $6$ isp $=\frac{1}{6}.$
$\Rightarrow\text{q}=\frac{5}{6}$
To find probability that on tenth throw $4^{th}$ six appears, in the first nine throw $3$ six should appear.
Required probability $= P(3$ six in first $9$ throw$) \times P($a six in tenth throw$)$
Required probability $=\text{ }^9\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^6\times\frac{1}{6}$
Required probability $=\frac{84\times5^6}{6^{10}}$

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MCQ 1541 Mark

If A and B are two events such that P(A) = 0.4, P(B) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ then $\text{P}(\overline{\text{B}}\cap\text{A})$ equals.

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

Answer

Correct option: D.

$\frac{1}{5}$

P(A) = 0.4, P(B) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5$

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

$\text{P}(\overline{\text{B}}\cap\text{A})=\text{P(A)}-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})\big]$

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

$\text{P}(\overline{\text{B}}\cap\text{A})=0.5-0.3$

$\text{P}(\overline{\text{B}}\cap\text{A})=0.2$

$\text{P}(\overline{\text{B}}\cap\text{A})=\frac{1}{5}$

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MCQ 1551 Mark

A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is

A
$\frac{1}{4}$
B
$\frac{11}{14}$
C
$\frac{15}{24}$
✓
$\frac{23}{24}$

Answer

Correct option: D.

$\frac{23}{24}$

4 letter can be placed in 4 envelopes in 4! ways = 24ways

Now, there is only one method, by which all the letters are placed in the right envelope.

P(all letters are placed in the envelopes) $=\frac{1}{24}$

P(all letters are not placed in the right envelopes) = 1 - P(all letters are placed in the right envelopes)

$=1-\frac{1}{24}=\frac{23}{24}$

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MCQ 1561 Mark

Choose the correct answer from the given four options.
The probability distribution of a discrete random variable X is given below:

$\text{X}$	$2$	$3$	$4$	$5$
$\text{P}(\text{X})$	$\frac{5}{\text{k}}$	$\frac{7}{\text{k}}$	$\frac{9}{\text{k}}$	$\frac{11}{\text{k}}$

The value of k is:

A
8.
B
16.
✓
32.
D
48.

Answer

Correct option: C.

32.

We know that, $\sum\text{P}\text{X}=1$

$\Rightarrow\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$

$\Rightarrow\frac{32}{\text{k}}=1$

$\therefore\text{k}=32$

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MCQ 1571 Mark

If $\text{P(A)}=\frac{3}{10},\text{P(B)}=\frac{2}{5}$ and $\text{P}(\text{A}\cap\text{B}=\frac{3}{5,}$ then P(A|B) + P(B|A) equals

A
$\frac{1}{4}$
✓
$\frac{7}{12}$
C
$\frac{5}{12}$
D
$\frac{1}{3}$

Answer

Correct option: B.

$\frac{7}{12}$

$\text{P(B)}=\frac{3}{10},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{5},\text{P}(\text{A}\cup\text{B})=\frac{3}{5}$

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

$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}$

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}+\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\frac{1}{10}}{\frac{2}{5}}+\frac{\frac{1}{10}}{\frac{3}{10}}$

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{7}{12}$

Note: Option is modified.

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MCQ 1581 Mark

In each of the following choose the correct answer:If A and B are events such that $\text{P}(\text{A}|\text{B})=\text{P}(\text{B}|\text{A}),\ \text{then}:$

A
$\text{A}\subset\text{B}\ \text{but}\ \text{A}\neq\text{B}$
B
$\text{A}=\text{B}$
C
$\text{A}\cap\text{B}=\phi$
✓
$\text{P}(\text{A})=\text{P}(\text{B})$

Answer

Correct option: D.

$\text{P}(\text{A})=\text{P}(\text{B})$

$\text{P}(\text{A}|\text{B})=\text{P}(\text{B}|\text{A})$ $\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$$\Rightarrow\ \text{P}(\text{A})=\text{P}(\text{B})$
Therefore, option (D) is correct.

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MCQ 1591 Mark

A bag contains 5 red and 3 blue balls are drawn at random without replacement, then the probability of getting exactly one red ball is.

A
$\frac{15}{29}$
✓
$\frac{15}{56}$
C
$\frac{45}{196}$
D
$\frac{135}{392}$

Answer

Correct option: B.

$\frac{15}{56}$

Total balls = 5 red + 3 blue = 8
Let R be the event of getting red ball
B be the event of getting a blue ball.
Required probability = P(BBR) + R(BRB) + P(RBB)
$=\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}+\frac{3}{8}\times\frac{5}{7}\times\frac{2}{6}+\frac{5}{8}\times\frac{3}{7}\times\frac{2}{6}$
$=\frac{15}{56}$

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MCQ 1601 Mark

If X follows a binomial distribution with parameter $\text{n}=8$ and $\text{p}=\frac{1}{2},$ then $\text{P(|X}-4|\leq2)$ equals:

A
$\frac{118}{128}$
✓
$\frac{119}{128}$
C
$\frac{117}{128}$
D
$\text{None of these}$

Answer

Correct option: B.

$\frac{119}{128}$

$\text{n = 8,}\text{p}=\frac{1}{2}=\text{q}$$\text{P(|X}-4|)\leq2$
$\Rightarrow-2\leq\text{x}-4\leq2$
$\Rightarrow4-2\leq\text{x}\leq2+4$
$\Rightarrow2\leq\text{x}\leq6$
$\text{P}(2\leq\text{x}\leq6)=\text{P(2)+P(3)+P(4)+P(5)+P(6)}$
$\text{P(2}\leq\text{x}\leq6)=\text{ }^8\text{C}_2\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_3\Big(\frac{1}{2^8}\Big)\\+\text{ }^8\text{C}_4\Big(\frac{1}{2^8}\Big)\text{ }^8\text{C}_5\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_6\Big(\frac{1}{2^8}\Big)$
$=\frac{119}{128}$

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MCQ 1611 Mark

Choose the correct answer from the given four options.
The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons 4 are swimmers is:

✓
${^5}\text{C}_4(0.7)^4(0.3)$
B
${^5}\text{C}_1(0.7)^4(0.3)^4$
C
${^5}\text{C}_4(0.7)(0.3)^4$
D
$(0.7)^4(0.3)$

Answer

Correct option: A.

${^5}\text{C}_4(0.7)^4(0.3)$

Here, $\bar{\text{p}}=0.3\Rightarrow\text{p}=0.7$
and $\text{q}=0.3,\text{n}=5$ and $\text{r}=4$
$\therefore$ Required probability $={^5}\text{C}_4(0.7)^4(0.3)$

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MCQ 1621 Mark

A coin is tossed 10 times. The probability of getting exactly six heads is:

A
$\frac{512}{513}$
✓
$\frac{105}{512}$
C
$\frac{100}{153}$
D
$\text{ }^{10}\text{C}_6$

Answer

Correct option: B.

$\frac{105}{512}$

$\text{n}=10,\text{x}=6,\text{p = q}=\frac{1}{2}$
$\text{P(X}=6)=\text{ }^{10}\text{C}_6\big(\frac{1}{2}\big)^{10}=\frac{105}{512}$

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MCQ 1631 Mark

Choose the correct answer from the given four options.
Two cards are drawn from a well shuffled deck of 52 playing cards with replacement. The probability, that both cards are queens, is:

✓
$\frac{1}{13}\times\frac{1}{13}$
B
$\frac{1}{13}\times\frac{1}{13}$
C
$\frac{1}{13}\times\frac{1}{17}$
D
$\frac{1}{13}\times\frac{4}{15}$

Answer

Correct option: A.

$\frac{1}{13}\times\frac{1}{13}$

Required probability $=\frac{4}{52}\cdot\frac{4}{52}$
$=\frac{1}{13}\times\frac{1}{13}$ [with replacement]

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MCQ 1641 Mark

Choose the correct answer from the given four options.If $\text{P}(\text{A})=0.4,\text{P}(\text{B})=0.8$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6,$ then $\text{P}(\text{A}\cup\text{B})$ is equal to:

A
0.24
B
0.3
C
0.48
✓
0.96

Answer

Correct option: D.

0.96

Here, P(A) = 0.4, P(B) = 0.8

and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6,$

$\because\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$

$\Rightarrow\text{P}(\text{B}\cap\text{A})=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)\cdot\text{P}(\text{A})$

$=0.6\times0.4=0.24$

$=1.2-0.24=0.96$

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MCQ 1651 Mark

A bag contains 12 balls out of which x are white. If one ball is drawn at random, what is the probability it will be a white ball?

A
$\frac{\text{x}}{2}$
✓
$\frac{\text{x}}{12}$
C
$\frac{\text{x}}{10}$
D
$\frac{12}{\text{x}}$

Answer

Correct option: B.

$\frac{\text{x}}{12}$

Total number of balls = 12

Number of white balls = x

P (white ball) $=\frac{\text{x}}{12}$

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MCQ 1661 Mark

Choose the correct answer from the given four options.
A and B are events such that P(A) = 0.4, P(B) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ Then $\text{P}(\text{B}'\cap\text{A})$ equals:

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

Answer

Correct option: D.

$\frac{1}{5}$

Here, $\text{P}(\text{A}) = 0.4,\text{P}(\text{B}) = 0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5$
$\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

$\Rightarrow\text{P}(\text{A}\cap\text{B})=0.4+0.3-0.5=0.2$

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

$=0.4-0.2=0.2=\frac{1}{5}$

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MCQ 1671 Mark

If A and B are two events such that $\text{A}\subset\text{B}$ and $\text{P}(\text{B})\neq0,$ then which of the following is correct?

A
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{B})}{\text{P}(\text{A})}$
B
$\text{P}(\text{A}|\text{B})\ <\text{P}(\text{A})$
✓
$\text{P}(\text{A}|\text{B})\ \geq\text{P}(\text{A})$
D
None of these.

Answer

Correct option: C.

$\text{P}(\text{A}|\text{B})\ \geq\text{P}(\text{A})$

$\text{A}\subset\text{B}\ \Rightarrow\ \ \text{A}\cap\text{B}=\text{A P}\ \ \text{and}\ \ \text{P}(\text{B})\neq0$
$\Rightarrow\ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{A})}{\text{P}(\text{B})}$
Since $\text{P}(\text{B})\neq\ 0$
$\therefore\ \frac{\text{P}(\text{A})}{\text{P}(\text{B})}<1\ \ \ \ \ \Rightarrow\ \text{P}(\text{A})<\text{P}(\text{B})\ \Rightarrow\ \text{P}(\text{A}|\text{B})\geq\text{P}(\text{A})$
Hence, option (C) is correct.

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MCQ 1681 Mark

If $\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13}.$ then, $\text{P}(\overline{\text{A}}|\text{B})=$

✓
$\frac{5}{9}$
B
$\frac{4}{9}$
C
$\frac{4}{13}$
D
$\frac{6}{13}$

Answer

Correct option: A.

$\frac{5}{9}$

We have,
$\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13}$
As, $\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{9}{13}-\frac{4}{13}$
$=\frac{5}{13}$
Now,
$\text{P}(\overline{\text{A}}|\text{B})=\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}$
$=\frac{\Big(\frac{5}{13}\Big)}{\Big(\frac{9}{13}\Big)}$
$=\frac{5}{9}$
Hence, the correct alternative is option $(a).$

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MCQ 1691 Mark

Two dice are thrown. If it is known that the sun of the numbers on the dice was less than 6, than the probability of gettinga sum 3, is

A
$\frac{1}{18}$
B
$\frac{5}{18}$
✓
$\frac{1}{5}$
D
$\frac{2}{5}$

Answer

Correct option: C.

$\frac{1}{5}$

$\text{S}=\begin{Bmatrix} (1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),\$2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6),\$3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6),\$4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6),\$5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6),\$6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6, 6) \end{Bmatrix}$

$\text{n(S)}=36$

Let A be the event that sum of the numbers on dice was less than 6.

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

$\text{n(A)} = 10$

Let B be the event that getting sum 3.

$\text{B}=\{(1, 2), (2, 1)\}\Rightarrow\text{n(B)}=2$

$\text{A}\cap\text{B}=\{(1,2),(2,1)\}\Rightarrow\text{n}(\text{A}\cap\text{B})=2$

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$

$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{10}=\frac{1}{5}$

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MCQ 1701 Mark

If $\text{P(A)}=\frac{2}{5},\text{P(B)}=\frac{3}{10}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5},$ then, $\text{P}(\overline{\text{A}}|\overline{\text{B}}) \text{ P}(\overline{\text{B}}|\overline{\text{A}})$ is equal to

A
$\frac{5}{6}$
B
$\frac{5}{7}$
✓
$\frac{25}{42}$
D
$1$

Answer

Correct option: C.

$\frac{25}{42}$

$\text{P(A)}=\frac{2}{5},\text{P(B)}=\frac{3}{10}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5}$

$\text{P}(\text{A}\cap\text{B})=\frac{1}{5}$

$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{1}{5}$

$\frac{2}{5}+\frac{3}{10}-\text{P}(\text{A}\cup\text{B})=\frac{1}{5}$

$\text{P}(\text{A}\cup\text{B})=\frac{1}{2}$

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

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\big[\text{P}(\overline{\text{A}\cup\text{B}})\big]^2}{\text{P}(\overline{\text{B}})\text{ P}(\overline{\text{A}})}$

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\big[1-\text{P}(\text{A}\cup\text{B})\big]^2}{\text{P}(\overline{\text{B}})\text{ P}(\overline{\text{A}})}$

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\Big[1-\frac{1}{2}\Big]^2}{\frac{7}{10}\times\frac{3}{5}}$

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{25}{42}$

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MCQ 1711 Mark

The probablity of selecting a male or a female is same. If the probability that in an office of n persons (n - 1) males being selected is $\frac{3}{2^{10}},$ the value of n is:

A
5
B
3
C
10
✓
12

Answer

Correct option: D.

X represents number of males.

$\text{p = q}=\frac{1}{2}$

$\text{p(n}-1)=\frac{3}{2^{10}}$

$\text{ }^{\text{n}}\text{C}_{\text{n}-1}\text{p}^{\text{n}-1}\text{q}=\frac{3}{2^{10}}$

$\text{n}\big(\frac{1}{2}\big)^{\text{n}-1}\big(\frac{1}{2}\big)^{\text{n}}=\frac{3}{2^{10}}$

$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=\frac{1}{4}\times\frac{3\times4}{2^{10}}$

$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=12\big(\frac{1}{2}\big)^{12}$

$\Rightarrow\text{n}=12$

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MATHS M.C.Q (1 Marks)