MCQ

MCQ 511 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 521 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 531 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 541 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 $= 24$ ways
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 551 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 561 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 $= \ce{P(BBR) + R(BRB) + P(RBB)}$
$=\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}+\frac{3}{8}\times\frac{5}{7}\times\frac{2}{6}+\frac{5}{8}\times\frac{3}{7}\times\frac{2}{6}$
$=\frac{15}{56}$

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MCQ 571 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 581 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 591 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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Maths MCQ