Maths M.C.Q (1 Marks)

4. $\frac{1}{5}$

Solution:

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

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

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

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

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

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

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

3. $\frac{17}{19}$

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 $=1-\frac{10}{20}\times\frac{9}{19}\times\frac{8}{18}=\frac{17}{19}$

2. $\frac{7}{12}$

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.

2. $\frac{15}{56}$

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)

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

$=\frac{15}{56}$

2. $\frac{5}{9}$

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).

4. $\frac{23}{24}$

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)

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

3. $\frac{1}{5}$

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.

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

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

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

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

3. $\frac{25}{42}$

Solution:

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

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

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

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

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

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

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

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

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

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

1. $\frac{10}{13}$

Solution:

Let E₁ 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 E₂ 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.

$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$

$\Rightarrow\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{13}$