M.C.Q (1 Marks)

MCQ 1511 Mark

The probability that in a year of 22^nd century chosen at random, there will be 53 Sunday, is

A
$\frac{3}{28}$
B
$\frac{2}{28}$
C
$\frac{7}{28}$
D
$\frac{5}{28}$

Answer

$\frac{5}{28}$

Solution:

We know a leap year is fallen within 4 years, So its probability is $\frac{25}{100}=\frac{1}{4}$

53rd Sunday leap year $=\frac{1}{4}\times\frac{2}{7}=\frac{2}{28}$

Similarly probability of 53rd Sunday in a non leap year $=\frac{75}{100}\times\frac{1}{7}=\frac{3}{4}\times\frac{1}{7}=\frac{3}{28}$

Required probability $=\frac{2}{28}+\frac{3}{28}=\frac{5}{28}$.

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MCQ 1521 Mark

Choose the correct answer from the given four options.
Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6, the probability of getting a sum 3, is:

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

Answer

$\frac{1}{5}$

Solution:

Let E₁ = Event that the sum of numbers on the dice was less than 6

And E₂= Event that the sum of numbers on the dice is 3.

$\therefore$ E₁ = {(1, 4), (4, 1), (2, 3), (3, 2), (2, 2), (1, 3), (3, 1), (1, 2), (2, 1), (1, 1)}

⇒ n(E₁) = 10

And E₂ = {(1, 2), (2, 1)}

⇒ n(E₂) = 2

$\therefore$ Required Probability $=\frac{2}{10}=\frac{1}{5}$

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MCQ 1531 Mark

A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 22 green balls and one blue ball is

A
$\frac{167}{168}$
B
$\frac{1}{28}$
C
$\frac{2}{21}$
D
$\frac{3}{28}$

Answer

$\frac{3}{28}$

Solution:

Total balls in a box - 3orange + 3green + 2blue = 8

Three balls are drawn at random from the box then samplw space $\text{n(S)}= {^{8}}\text{C}_3=\frac{8\times7\times6}{3\times2\times1}=56$

Let A be the event that drawing 2 green and one blue ball.

$\text{n(A)}={^{3}}\text{C}_2\times{^{2}}\text{C}_2=6$

$\text{P(A)}=\frac{6}{56}=\frac{3}{28}$

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MCQ 1541 Mark

A bouquet from 11 different flowers is to be made so that it contains not less then three flowers. Then the number of the different ways of selecting flowers to from the bouquet.

A
1972
B
1952
C
1981
D
1947

Answer

1972

Solution:

No. of ways = ¹¹C₃ + ¹¹C₄ + ¹¹C₅ + ¹¹C₆ + ¹¹C₇ + ¹¹C₈ + ¹¹C₉ + ¹¹C₁₀ + ¹¹C₁₁

⇒ 165 + 330 + 462 + 462 + 330 + 165 + 55 + 11 + 1

⇒ 1981

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MCQ 1551 Mark

A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one iten is chosen ar random, the probability that it is rusted or is nail is

A
$\frac{3}{16}$
B
$\frac{5}{16}$
C
$\frac{11}{16}$
D
$\frac{14}{16}$

Answer

$\frac{11}{16}$

Solution:

Rusted items = 3 + 5 = 8

Rusted nails = 3

Total nails = 6

P(getting a rusted item or a nail) = P(getting a rusted item) + P(getting a nail) - P(getting a rusted item and a nail)

$=\frac{8}{16}+\frac{6}{16}-\frac{3}{16}$

$=\frac{8+6-3}{16}$

$=\frac{11}{16}$

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MCQ 1561 Mark

Let X denote the number of times heads occur in n tosses of a fair coin. If P(X = 4), P(X = 5) and P(X = 6) are in AP, the value of n is:

A
7, 14
B
10, 14
C
12, 7
D
14, 12

Answer

7, 14

Solution:

X denotes the number of times heads occurs.

P(X = 4),P(X = 5),P(X = 6) are in AP

$\Rightarrow2\text{P(X = 4),P(X = 5),P(X = 6)}$

$\Rightarrow2\text{ }^{\text{n}}\text{C}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^{\text{n}-5}=\text{ }^{\text{n}}\text{C}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{\text{n}-4}\times\text{ }^{\text{n}}\text{C}_6\big(\frac{1}{2}\big)^6\big(\frac{1}{2}\big)^{\text{n}-6}$

$\Rightarrow2\text{ }^{\text{n}}\text{C}_5\big(\frac{1}{2}\big)^{\text{n}}=\text{ }^{\text{n}}\text{C}_4\big(\frac{1}{2}\big)^{\text{n}}+\text{ }^{\text{n}}\text{C}_6\big(\frac{1}{2}\big)^{\text{n}}$

$\Rightarrow2\text{ }^{\text{n}}\text{C}_5=\text{ }^{\text{n}}\text{C}_4+\text{ }^{\text{n}}\text{C}_6$

$\Rightarrow\frac{2\text{n}!}{5!(\text{n}-5)!}=\frac{\text{n}!}{4!(\text{n}-4)!}+\frac{\text{n}!}{6!(\text{n}-6)!}$

$\Rightarrow\frac{2}{5\times4!(\text{n}-5)(\text{n}-6)!}=\frac{1}{4!(\text{n}-4)(\text{n}-5)(\text{n}-6)!}+\frac{1}{6\times5\times4!(\text{n}-6)!}$

$\Rightarrow\frac{2}{5(\text{n}-5)}=\frac{1}{(\text{n}-4)(\text{n}-5)}+\frac{1}{6\times5}$

$\Rightarrow\frac{2}{5(\text{n}-5)}=\frac{30+(\text{n}-4)(\text{n}-5)}{30(\text{n}-4)(\text{n}-5)}$

$\Rightarrow12(\text{n}-4)=30+(\text{n}-4)(\text{n}-5)$

$\Rightarrow12(\text{n}-4)-(\text{n}-4)(\text{n}-5)=30$

$\Rightarrow(\text{n}-4)(12-\text{n}+5)=30$

$\Rightarrow(\text{n}-4)(17-\text{n})=30$

Check with options by putting value of n.

$\Rightarrow\text{n}=7,14$

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MCQ 1571 Mark

If A and B are two events associated to a random experiment such that $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P(B)}=\frac{17}{20}$, then P(A|B) =

A
$\frac{14}{17}$
B
$\frac{17}{20}$
C
$\frac{7}{8}$
D
$\frac{1}{8}$

Answer

$\frac{14}{17}$

Solution:

$\text{P}(\text{A}\cap\text{B})=-\frac{7}{10},\text{P(B)}=\frac{17}{20}$

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

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{7}{10}}{\frac{17}{20}}=\frac{14}{17}$

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MCQ 1581 Mark

A flash light has 8 batteries out of which 3 are dead. IF two batteries are selected without replacement and tested, then the probability that both are dead is,

A
$\frac{3}{28}$
B
$\frac{1}{14}$
C
$\frac{9}{64}$
D
$\frac{33}{56}$

Answer

$\frac{3}{28}$

Solution:

We have,

The total number of batteries = 8

The number of dead batteries = 3

Let A be the event of selecting the first dead battery and B be the event of selecting the second dead battery.

Now,

P(both dead batteries are selected) $=\text{P}(\text{A}\cap\text{B})$

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

$=\frac{3}{8}\times\frac{2}{7}$

$=\frac{3}{28}$

Hence, the correct alternative is option (a).

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MCQ 1591 Mark

A special lottery is to be held to select a student who will live in the only deluxe room in a hostel. There are 100 Year-III, 150 Year-II and 200 Year-I students who applied. Each Year-III's name is placed in the lottery 3 times; each Year-II's name, 2 times and Year-I's name, 1 time. What is the probability that a Year-III's name will be chosen?

A
$\frac18$
B
$\frac28$
C
$\frac38$
D
$\frac12$

Answer

$\frac38$

Solution:

Total names in the lottery

= 3 × 100 + 2 × 150 + 200 = 800

Number of Year-III's names = 3 × 100 = 300

Required probability $=\frac{300}{800}=\frac38$

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MCQ 1601 Mark

If A and B are two independent events with $\text{P(A)}=\frac{3}{5}$ and $\text{P(B)}=\frac{4}{9},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals,

A
$\frac{4}{15}$
B
$\frac{8}{45}$
C
$\frac{1}{3}$
D
$\frac{2}{9}$

Answer

$\frac{2}{9}$

Solution:

$\text{P(A)}=\frac{3}{5},\text{P(B)}=\frac{4}{9}$

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

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

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{3}{5}+\frac{4}{9}-\frac{3}{5}\times\frac{4}{9}\Big]$

($\because$ A and B are independent)

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\frac{7}{9}$

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{2}{9}$

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MCQ 1611 Mark

The probability that a leap year will have 53 fridays or 53 Saturdays is.

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

Answer

$\frac{3}{7}$

Soluction:

Non-leap year has 365 days = 52 weeks + 1

366 days in leap year.

We want to find probability of 53 Fridays or 53 Saturday.

Favourable cases = {(Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}

Required probability $=\frac{3}{7}$

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MCQ 1621 Mark

Choose the correct answer from the given four options:
If A and B are such events that $\text{P}(\text{A})>0$ and $\text{P}(\text{B})\neq1,$ then $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)$ equals to:

A
$1-\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
B
$1-\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)$
C
$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B}')}$
D
$\frac{\text{P}(\text{A}')}{\text{P}(\text{B}')}$

Answer

$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B}')}$

Solution:

We have, $\text{P}(\text{A})>0$ and $\text{P}(\text{B})\neq1$

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

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

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MCQ 1631 Mark

Choose the correct answer from the given four options.
Let A and B be two events such that $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}.$Then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:

A
$\frac{2}{5}$
B
$\frac{3}{8}$
C
$\frac{3}{20}$
D
$\frac{6}{25}$

Answer

$\frac{6}{25}$

Solution:

We have, $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}$

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

$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{3}{8}+\frac{5}{8}-\frac{3}{4}=\frac{1}{4}$

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

$=\frac{\frac{1}{4}}{\frac{5}{8}}=\frac{2}{5}$

and $\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)=\frac{\text{P}(\text{A}'\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$

$=\frac{\frac{5}{8}-\frac{1}{4}}{\frac{5}{8}}=\frac{3}{5}$

$\therefore\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)$

$=\frac{2}{5}\cdot\frac{3}{5}=\frac{6}{25}$

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MCQ 1641 Mark

Choose the correct answer from the given four options.
A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is:

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

Answer

$\frac{15}{56}$

Solution:

Probability of getting exactly one red (R) ball

$=\text{P}_{\text{R}}\cdot\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\bar{\text{R}}}+\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\text{R}}\cdot\text{P}_{\bar{\text{R}}}+\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\text{R}}$

$=\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}+\frac{3}{8}\cdot\frac{5}{7}\cdot\frac{2}{7}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{5}{6}$

$=\frac{15}{4\cdot7\cdot6}+\frac{15}{4\cdot7\cdot6}+\frac{15}{4\cdot7\cdot6}$

$=\frac{5}{56}+\frac{5}{56}+\frac{5}{56}=\frac{15}{56}$

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MCQ 1651 Mark

Two cards are drawn from a well shuffled deck of 52 playing cards with replacement. The probability that both cards are queen is

A
$\frac{1}{13}\times\frac{1}{13}$
B
$\frac{1}{13}+\frac{1}{13}$
C
$\frac{1}{13}\times\frac{1}{17}$
D
$\frac{1}{13}\times\frac{4}{5}$

Answer

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

Solution:

Two cards are drawn from 52 cards.

Let, E₁ be the event that getting queen in first draw and E₂ be the event that getting queen in second draw,

$\text{P}(\text{E}_1\cap\text{E}_2)=\frac{4}{52}\times\frac{4}{52}=\frac{1}{13}\times\frac{1}{13}$

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MCQ 1661 Mark

Choose the correct answer from the given four options.
The probability that exactly two of the three balls were red, the first ball being red, is:

A
$\frac{1}{3}$
B
$\frac{4}{7}$
C
$\frac{15}{28}$
D
$\frac{5}{28}$

Answer

$\frac{4}{7}$

Solution:

Let E₁= Event that first ball being red

And E₂ = Event that exactly two of three balls being red

$\therefore\text{P}(\text{E}_1)=\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\bar{\text{R}}}$

$=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}$

$=\frac{60+60+60+30}{336}=\frac{210}{336}$

$\text{P}(\text{E}_1\cap\text{E}_2)=\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\text{R}}+\text{P}{_\text{R}}\cdot\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}$

$=\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{120}{336}$

$\therefore\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E}_2)}{\text{P}(\text{E}_1)}$

$=\frac{\frac{120}{336}}{\frac{210}{336}}=\frac{4}{7}$

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MCQ 1671 Mark

Choose the correct answer from the given four options.
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$ is equal to:

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

Answer

$\frac{3}{5}$

Solution:

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

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

$=\frac{\frac{3}{5}-\frac{3}{10}}{1-\frac{1}{2}}=\frac{\frac{6-3}{10}}{\frac{1}{2}}$

$=\frac{6}{10}=\frac{3}{5}$

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MCQ 1681 Mark

If one ball is drawn ar random from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, then the probability that 2 white and 1 black balls will be drawn is.

A
$\frac{13}{32}$
B
$\frac{1}{4}$
C
$\frac{1}{32}$
D
$\frac{3}{16}$

Answer

$\frac{13}{32}$

Solution:

Total balls in first box = 3 white + 1 black = 4

Total balls in second box = 2 white + 2black = 4

Total balls in third box = 1white + 3black = 4

Probability of 2 white and 1 black

= P(WWB) + P(WBW) + P(BWW)

$=\frac{3}{4}\times\frac{2}{4}\times\frac{3}{4}+\frac{3}{4}\times\frac{2}{4}\times\frac{1}{4}+\frac{1}{4}\times\frac{2}{4}\times\frac{1}{4}$

$=\frac{18+6+2}{64}=\frac{13}{32}$

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MCQ 1691 Mark

From a set of 100 cards numbered 1 to 100, one card is drawn at randow. The probability number obtained on the card is divisible by 6 or 8 but not by 24 is

A
$\frac{6}{25}$
B
$\frac{1}{4}$
C
$\frac{1}{6}$
D
$\frac{2}{6}$

Answer

$\frac{6}{25}$

Solution:

Number of cards divisible by 6 = 16

$\Rightarrow\ \text{P(A)}=\frac{16}{100}$

Number of cards divisible by 8 = 12

$\Rightarrow\ \text{P(B)}=\frac{12}{100}$

Number of cards divisible by 24 = 4

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{4}{100}$

$\text{P}(\text{A}\cup\text{B})=\frac{16}{100}+\frac{12}{100}-\frac{4}{100}$

$\text{P}(\text{A}\cup\text{B})=\frac{6}{25}$

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MCQ 1701 Mark

Choose the correct answer from the given four options.
If $\text{P}(\text{A})=\frac{3}{10},\text{P}(\text{B})=\frac{2}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{5},$ then $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)+\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ equas:

A
$\frac{1}{4}$
B
$\frac{1}{3}$
C
$\frac{15}{12}$
D
$\frac{7}{2}$

Answer

$\frac{7}{12}$

Solution:

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

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

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

$\therefore\text{P}(\text{A}\cap\text{B})=\frac{1}{10}$

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

$=\frac{\frac{1}{10}}{\frac{3}{10}}+\frac{\frac{1}{10}}{\frac{2}{5}}$

$=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$

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MCQ 1711 Mark

If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{B}|\overline{\text{A}})=$

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

Answer

$\frac{3}{5}$

Solution:

$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$

Consider,

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

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

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

$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}$

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

$\text{P(A)}+\frac{3}{5}-\frac{4}{5}=\frac{3}{10}$

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

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

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

$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{3}{5}$

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MCQ 1721 Mark

If S is the samle space and $\text{P(A)}=\frac{1}{3}, \text{P(B)}$ and $\text{S}=\text{A}\cup\text{B,}$ where A and B are tow mutually exclusive events, then P(A) =

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

Answer

$\frac{1}{4}$

Solution:

$\text{P(A)}=\frac{1}{3}\text{P(A)}$

$\Rightarrow\ \text{P(B)}=3\text{P(A)}\ .....(\text{i})$

A and B are mutually exclusive events.

$\Rightarrow\ \text{P}(\text{A}\cap\text{B}) = 0$

Now,

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

$\Rightarrow\ \text{P(A)}+\text{P(B)}=1$

$\Rightarrow\ \text{P(A)}+3\text{P(A)}=1$ [From (i)]

$\Rightarrow\ 4\text{P(A)}=1$

$\Rightarrow\ \text{P(A)}=\frac{1}{4}$

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MCQ 1731 Mark

Choose the correct answer from the given four options.
If $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{A}\cup\text{B})'+\text{P}(\text{A}'\cup\text{B})=$

A
$\frac{1}{5}$
B
$\frac{4}{5}$
C
$\frac{1}{2}$
D
$1.$

Answer

$1.$

Solution:

We have, $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$

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

$=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$

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

$\Rightarrow\text{P}(\text{A})=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{2}$

$\therefore\text{P}(\text{A}\cup\text{B})'=1-\text{P}(\text{A}\cup\text{B})$

$=1-\frac{4}{5}=\frac{1}{5}$

And $\text{P}(\text{A}'\cap\text{B})=1-\text{P}(\text{A}-\text{B})$

$=1-\big[\text{P}(\text{A})-\text{P}(\text{A}\cap\text{B})\big]$

$=1-\Big(\frac{1}{2}-\frac{3}{10}\Big)=\frac{4}{5}$

$\Rightarrow\text{P}(\text{A}\cup\text{B})'+\text{P}(\text{A}'\cup\text{B})$

$=\frac{1}{5}+\frac{4}{5}=1$

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MCQ 1741 Mark

Let A and B be two events. If $\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$ then P(A|B) is equal to

A
0.8
B
0.5
C
0.3
D
0

Answer

Solution:

$\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$

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

$\Rightarrow \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}$

$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2+0.4-0.6}{\text{P(B)}}$

$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0$

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MCQ 1751 Mark

A bag contains six red four green and eight white balls If a ball is picked at random the probability that it is not white is:

A
$\frac13$
B
$\frac49$
C
$\frac59$
D
$\frac23$

Answer

$\frac59$

Solution:

Number of balls that are not white = 10

Total $=\frac 18$

$∴ $ P(not white) $= \frac{18}{10} = 95$

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MCQ 1761 Mark

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

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

Answer

$\frac{1}{36}$

Solution:

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 1771 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
$\frac{15}{29}$
B
$\frac{15}{56}$
C
$\frac{45}{196}$
D
$\frac{135}{392}$

Answer

$\frac{15}{56}$

Solution:

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 1781 Mark

Choose the correct answer in each of the following:
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is

A
$\frac{37}{221}$
B
$\frac{5}{13}$
C
$\frac{1}{13}$
D
$\frac{2}{13}$

Answer

$\text{n}(\text{S})=52,\ \text{n}(\text{A})=4$

$\text{P}(\text{X}=0)=\frac{^{48}\text{C}_2}{^{52}\text{C}_2}=\frac{48\times47}{52\times51}=\frac{188}{221}$

$\text{P}(\text{X}=1)=\frac{^{48}\text{C}_2\times^4\text{C}_1}{^{52}\text{C}_2}=\frac{2\times48\times4}{52\times51}=\frac{32}{221}$

$\text{P}(\text{X}=2)=\frac{^4\text{C}_2}{^{52}\text{C}_2}=\frac{4\times3}{52\times51}=\frac{1}{221}$

$\text{x}_i$

$\text{p}_i$

$\text{p}_i\text{x}_i$

$0$

$1$

$2$

$\frac{188}{221}$

$\frac{32}{221}$

$\frac{1}{221}$

$0$

$\frac{32}{221}$

$\frac{2}{221}$

$\sum\text{p}_i\text{x}_i=\frac{34}{221}=\frac{2}{13}$

$\text{Now}\ \text{E}(\text{X})=\frac{2}{13}$

Therefore, option (D) is correct.

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MCQ 1791 Mark

Choose the correct answer in each of the following:
If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then

A
$\text{A}\subset\text{B}$
B
$\text{B}\subset\text{A}$
C
$\text{B}=\phi$
D
$\text{A}=\phi$

Answer

$\text{A}\subset\text{B}$
$\text{P}(\text{B|A})=1$
$\Rightarrow\ \frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}=1\ \ \text{P}(\text{B}\cap\text{A})=\text{P}(\text{A})$
$\therefore$ (A) is correct answer.

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MCQ 1801 Mark

Probability that A speaks truth is $\frac{4}{5}.$ A coin is tossed. A reports that a head appears. The probability that actually there was head is

A
$\frac{4}{5}$
B
$\frac{1}{2}$
C
$\frac{1}{5}$
D
$\frac{2}{5}$

Answer

Let A be the event that the man reports that head occurs in tossing a coin and let E₁be the event that head occurs and E₂be the event head does not occur.
$\text{P}(\text{E}_1)=\frac{1}{2},\ \text{P}(\text{E}_2)=\frac{1}{2}$
$\text{P}(\text{A}|\text{E}_1)$ = P(A reports that head occurs when head had actually occur red on the coin) = $\frac{4}{5}$
$\text{P}(\text{A}|\text{E}_2)=$ P(A reports that head occurs when head had not occur red on the coin) $=1-\frac{4}{5}=\frac{1}{5}$
By Bayes’ theorem,
$ \text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+{\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)}}=\frac{\frac{1}{2}\times\frac{4}{5}}{\frac{1}{2}\times\frac{4}{5}+\frac{1}{2}\times\frac{1}{5}}=\frac{4}{4+1}=\frac{4}{5}$
Hence, option (A) is correct.

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MCQ 1811 Mark

In each of the following, choose the correct answer:
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

A
$10^{-1}$
B
$\Big(\frac{1}{2}\Big)^5$
C
$\Big(\frac{9}{10}\Big)^5$
D
$\frac{9}{10}$

Answer

The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.
Probability of getting a defective bulb, p $=\frac{10}{100}=\frac{1}{10}$
$\therefore\text{q}=1-\text{p}=1-\frac{1}{10}=\frac{9}{10}$
Clearly, X has a binomial distribution with n = 5 and $\text{p}=\frac{1}{10}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}=\ ^5\text{C}_\text{x}\bigg(\frac{9}{10}\bigg)^{5-\text{x}}\bigg(\frac{1}{10}\bigg)^\text{x}$
P(none of the bulbs is defective) = P(X = 0)
$=\ ^5\text{C}_0\cdot\Big(\frac{9}{10}\Big)^5$
$=1\cdot\Big(\frac{9}{10}\Big)^5$
$=\Big(\frac{9}{10}\Big)^5$
The correct answer is C.

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MCQ 1821 Mark

Choose the correct answer in each of the following:
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then

A
P(B|A) = 1
B
P(A|B) = 1
C
P(B|A) = 0
D
P(A|B) = 0

Answer

$\text{P}(\text{A|B})=1$
$\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A and B})=\text{P}(\text{A})$
$\Rightarrow\ \text{P}(\text{B})=\text{P}(\text{A}\cap\text{B})\ \Rightarrow\ 1=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=1=\text{P}(\text{A|B})$
$\therefore$ (B) is correct answer.

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MCQ 1831 Mark

Choose the correct answer in each of the following:
If P(A|B) > P(A), then which of the following is correct :

A
P(B|A) < P(B)
B
P(A ∩ B) < P(A).P(B)
C
P(B|A) > P(B)
D
P(B|A) = P(B)

Answer

$\text{P}(\text{B|A})>\text{P}(\text{B})$
$\text{P}(\text{A|B})>\text{P}(\text{A})$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}>\text{P}(\text{A})\ \ \Rightarrow\ \ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\text{P}(\text{B})$
$\Rightarrow\ {\text{P}(\text{B}|\text{A})}>{\text{P}(\text{B})}.$
(C) is correct answer.

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MCQ 1841 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
$\text{P}(\text{A}|\text{B})\ <\text{P}(\text{A})$
C
$\text{P}(\text{A}|\text{B})\ \geq\text{P}(\text{A})$
D
None of these.

Answer

$\text{A}\subset\text{B}\ \Rightarrow\ \ \text{A}\cap\text{B}=\text{A P}\ \ \text{and}\ \ \text{P}(\text{B})\neq0$
$\Rightarrow\ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{A})}{\text{P}(\text{B})}$
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 1851 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
$\text{A}\cap\text{B}=\phi$
D
$\text{P}(\text{A})=\text{P}(\text{B})$

Answer

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

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

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

Therefore, option (D) is correct.

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MCQ 1861 Mark

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

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

Answer

Correct option: D.

$\frac{7}{12}$

(d) : Given, $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}, P(A \cup B)=\frac{3}{5}$
Now, $P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$
=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}=\frac{3+4-6}{10}=\frac{1}{10}
$
Now, $P(B \mid A)+P(A \mid B)=\frac{P(A \cap B)}{P(A)}+\frac{P(A \cap B)}{P(B)}$
$
=\frac{1 / 10}{3 / 10}+\frac{1 / 10}{2 / 5}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}
$

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MCQ 1871 Mark

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. Then the probability of getting both ball red is

A
$\frac{12}{49}$
✓
$\frac{16}{49}$
C
$\frac{2}{7}$
D
None of these

Answer

Correct option: B.

$\frac{16}{49}$

(b) : Required probability $=\frac{4}{7} \times \frac{4}{7}=\frac{16}{49}$

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MCQ 1881 Mark

If $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$, then find the value of $P(B / A)$.

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

Answer

Correct option: B.

$\frac{1}{3}$

(b) : Given, $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}, P(A \cup B)=\frac{3}{5}$
Now, $P(A \cap B)=P(A)+P(B)-P(A \cup B)$
$
=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}=\frac{3+4-6}{10}=\frac{1}{10}
$
Now, $P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{1 / 10}{3 / 10}=\frac{1}{3}$

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MCQ 1891 Mark

If $P(A)=\frac{7}{13}, P(B)=\frac{9}{13}$ and $P(A \cup B)=\frac{12}{13}$, then evaluate $P(A \mid B)$.

A
$\frac{2}{3}$
B
$\frac{4}{9}$
C
$\frac{4}{5}$
D
None of these

Answer

$
\begin{array}{l}
\text { (b) : Given, } P(A)=\frac{7}{13}, P(B)=\frac{9}{13} \text { and } P(A \cup B)=\frac{12}{13} \\
\therefore \quad P(A \cap B)=P(A)+P(B)-P(A \cup B) \\
=\frac{7}{13}+\frac{9}{13}-\frac{12}{13} \Rightarrow P(A \cap B)=\frac{4}{13} \\
\therefore \quad P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{4 / 13}{9 / 13}=\frac{4}{9}
\end{array}
$

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MCQ 1901 Mark

A bag contains 10 white and 6 black balls. 4 balls are successively drawn out without replacement. What is the probability that they are alternately of different colours?

✓
0.12
B
0.08
C
0.11
D
0.07

Answer

Correct option: A.

0.12

(a) : Required probability $=P(B W B W)+P(W B W B)$
$
=\frac{6}{16} \cdot \frac{10}{15} \cdot \frac{5}{14} \cdot \frac{9}{13}+\frac{10}{16} \cdot \frac{6}{15} \cdot \frac{9}{14} \cdot \frac{5}{13}=\frac{90}{728}=\frac{45}{364} \text {. }
$

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MCQ 1911 Mark

Suppose that five good fuses and two defective ones have been mixed up. To find the defective fuses, we test them one-by-one, at random and without replacement. What is the probability that we are lucky and find both of the defective fuses in the first two tests?

A
$\frac{1}{42}$
B
$\frac{2}{21}$
C
$\frac{1}{18}$
✓
$\frac{1}{21}$

Answer

Correct option: D.

$\frac{1}{21}$

(d) : Let $D_1, D_2$ be the events that we find a defective fuse in the first and second test respectively.
$\therefore \quad$ Required probability $=P\left(D_1 \cap D_2\right)$
$
=P\left(D_1\right) P\left(D_2 \mid D_1\right)=\frac{2}{7} \cdot \frac{1}{6}=\frac{1}{21}
$

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MCQ 1921 Mark

$A$ and $B$ are events such that $P(A)=0.4, P(B)$ $=0.3$ and $P(A \cup B)=0.5$. Then $P\left(B^{\prime} \cap A\right)$ equals

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

Answer

Correct option: D.

$\frac{1}{5}$

(d) : $P(A)=0.4, P(B)=0.3, P(A \cup B)=0.5$
$P(A \cap B)=P(A)+P(B)-P(A \cup B)=0.4+0.3-0.5=0.2$
Now, $P\left(B^{\prime} \cap A\right)=P(A)-P(A \cap B)=0.4-0.2=0.2=\frac{1}{5}$

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MCQ 1931 Mark

If six cards are selected at random (without replacement) from a standard deck of 52 cards, what is the probability that there will be no pairs? (two cards of same denomination)

A
0.28
B
0.562
✓
0.345
D
0.832

Answer

Correct option: C.

0.345

(c) : Let $E_i$ be the event that the first $i$ cards have no pair among them. Then we want to compute $P\left(E_6\right)$, which is actually the same as $P\left(E_1 \cap E_2 \cap \ldots \cap E_6\right)$, since $E_6 \subset E_5 \subset \ldots \subset E_1$, implying that $E_1 \cap E_2 \cap \ldots \cap E_6=E_6$.
$
\begin{array}{l}
\therefore \quad P\left(E_1 \cap E_2 \cap \ldots \cap E_6\right)=P\left(E_1\right) P\left(E_2 \mid E_1\right) P\left(E_3 \mid E_1 \cap E_2\right) \ldots \\
=\frac{52}{52} \cdot \frac{48}{51} \cdot \frac{44}{50} \cdot \frac{40}{49} \cdot \frac{36}{48} \cdot \frac{32}{47}=0.345
\end{array}
$

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MCQ 1941 Mark

A random variable $X$ has the following distribution.

$X$	1	2	3	4	5	6	7	8
$P(X)$	0.15	0.23	0.12	0.10	0.20	0.08	0.07	0.05

For the event $E=\{X$ is prime number $\}$, find $P(E)$.

A
0.87
✓
0.62
C
0.35
D
0.5

Answer

Correct option: B.

0.62

(b) : $P(E)=P(X=2)+P(X=3)+P(X=5)+P(X=7)$
$=0.23+0.12+0.20+0.07=0.62$

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MCQ 1951 Mark

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

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

Answer

Correct option: C.

$\frac{15}{56}$

(c) : Possible outcomes $=\{(R B B),(B R B),(B B R)\}$
Required probability $=P(R B B)+P(B R B)+P(B B R)$
$
=\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6}=3 \times \frac{5}{56}=\frac{15}{56}
$

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MCQ 1961 Mark

If two events are independent, then

A
they must be mutually exclusive.
B
the sum of their probabilities must be equal to 1 .
C
both (a) and (b) are correct
✓
None of these

Answer

Correct option: D.

None of these

(d)

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MCQ 1971 Mark

If $A$ and $B$ are two independent events with $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$, then $P\left(B^{\prime} \mid A\right)$ is equal to

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

Answer

Correct option: C.

$\frac{3}{4}$

(c) : Given, $A$ and $B$ are independent events.
Also, $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$
Now, $P\left(B^{\prime} \mid A\right)=\frac{P\left(B^{\prime} \cap A\right)}{P(A)}=\frac{P\left(B^{\prime}\right) P(A)}{P(A)}$
$[\because A$ and $B$ are independent events $]$
$
=P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{4}=\frac{3}{4}
$

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MCQ 1981 Mark

If $P(A)=0.4, P(B)=0.8$ and $P(B \mid A)=0.6$, then $P(A \cup B)$ is equal to

A
0.24
B
0.3
C
0.48
✓
0.96

Answer

Correct option: D.

0.96

(d) : Given, $P(A)=0.4, P(B)=0.8$ and $P(B \mid A)=0.6$.
Clearly, $P(A \cap B)=P(B \mid A) P(A)=0.6 \times 0.4=0.24$
Now, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$=0.4+0.8-0.24=0.96$

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MCQ 1991 Mark

A random variable $X$ has the following probability distribution :

$X$	0	1	2	3	4	5	6	7
$P(X)$	$a$	$6 a$	$6 a$	$4 a$	$8 a$	$8 a$	$6 a$	$9 a$

Find the value of $a$.

A
$\frac{1}{47}$
✓
$\frac{1}{48}$
C
$\frac{1}{33}$
D
$\frac{1}{29}$

Answer

Correct option: B.

$\frac{1}{48}$

(b) : We know that, $\Sigma P_i=1$
$
\begin{array}{l}
\Rightarrow a+6 a+6 a+4 a+8 a+8 a+6 a+9 a=1 \\
\Rightarrow 48 a=1 \Rightarrow a=\frac{1}{48}
\end{array}
$

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MCQ 2001 Mark

The probability distribution of a discrete random variable $X$ is given below :

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

The value of $k$ is

A
8
B
16
✓
32
D
48

Answer

Correct option: C.

(c) : We have, $\Sigma P(X)=1$
$
\Rightarrow \frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k}=1 \Rightarrow \frac{32}{k}=1 \Rightarrow k=32
$

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Maths M.C.Q (1 Marks)