M.C.Q (1 Marks)

Question 511 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.

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

Answer

$\frac{1}{5}$

Solution:
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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Question 521 Mark

Three integers are chosen at random from the first 20 integers. The probability that their product is even is,

$\frac{2}{19}$
$\frac{3}{29}$
$\frac{17}{19}$
$\frac{4}{19}$

Answer

$\frac{17}{19}$

Soluction:
Required probability that product of two integers should be even.
10 integers are odd out of first 20 integers.
Required probability = 1 - Probability of product is odd
Product of three integers is odd if two numbers are odd
Required probability $=1-\frac{10}{20}\times\frac{9}{19}\times\frac{8}{18}=\frac{17}{19}$

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Question 531 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

$\frac{1}{4}$
$\frac{7}{12}$
$\frac{5}{12}$
$\frac{1}{3}$

Answer

$\frac{7}{12}$

Solution:
$\text{P(B)}=\frac{3}{10},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{5},\text{P}(\text{A}\cup\text{B})=\frac{3}{5}$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{A})$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}+\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\frac{1}{10}}{\frac{2}{5}}+\frac{\frac{1}{10}}{\frac{3}{10}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{7}{12}$
Note: Option is modified.

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Question 541 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.

$\frac{15}{29}$
$\frac{15}{56}$
$\frac{45}{196}$
$\frac{135}{392}$

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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Question 551 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, $\text{P}(\overline{\text{A}}|\text{B})=$

$\frac{5}{9}$
$\frac{4}{9}$
$\frac{4}{13}$
$\frac{6}{13}$

Answer

$\frac{5}{9}$

Solution:
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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Question 561 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

$\frac{1}{4}$
$\frac{11}{14}$
$\frac{15}{24}$
$\frac{23}{24}$

Answer

$\frac{23}{24}$

Soluction:
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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Question 571 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

$\frac{1}{18}$
$\frac{5}{18}$
$\frac{1}{5}$
$\frac{2}{5}$

Answer

$\frac{1}{5}$

Solution:
$\text{S}=\begin{Bmatrix} (1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),\$2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6),\$3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6),\$4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6),\$5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6),\$6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6, 6) \end{Bmatrix}$
$\text{n(S)}=36$
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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Question 581 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

$\frac{5}{6}$
$\frac{5}{7}$
$\frac{25}{42}$
$1$

Answer

$\frac{25}{42}$

Solution:
$\text{P(A)}=\frac{2}{5},\text{P(B)}=\frac{3}{10}$ 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 591 Mark

$A$ and $B$ are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability of their making common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is.

✓
$\frac{10}{13}$
B
$\frac{13}{120}$
C
$\frac{1}{40}$
D
$\frac{1}{12}$

Answer

Correct option: A.

$\frac{10}{13}$

Let $E_1$ be the event that Both $A$ and $B$ solve the problem.$A$ and $B$ are independent,
$\Rightarrow\ \text{P}(\text{E}_1)=\text{P(A)}\times\text{P(B)}$
$\Rightarrow\ \text{P}(\text{E}_1)=\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$
Let $E_2$ both $A$ and $B$ got wrong solution.
$\text{P}(\text{E}_2)=\Big(1-\frac{1}{3}\Big)\times\Big(1-\frac{1}{4}\Big)=\frac{1}{2}$
Let $E$ be the event getting same answer.
$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$
$\Rightarrow\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{13}$

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MATHS M.C.Q (1 Marks)