M.C.Q (1 Marks)

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

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

Answer

Correct option: B.

$\frac{4}{9}$

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 M.C.Q (1 Marks)