MATHS M.C.Q (1 Marks)

1. $\frac{1}{7}$

Solution:

There are 365 days in non-leap year and there are 7 days in a week

$\therefore\ 365\div7=52$ weeks + 1 days

So, this day may be Tuesday or Wednesday.

So, the required probability $=\frac{1}{7}$

3. $\frac{1}{132}$

Solution:

If all the girls sit together, then we consider it as 1 group

$\therefore$ Total number of arrangement of 6 + 1 = 7 persons in a row = 7! And the girls also interchanged their places with 6! Ways.

$\therefore\ \text{Required probability}=\frac{6!7!}{12!}$

$=\frac{6\times5\times4\times3\times2\times7!}{12\times11\times10\times9\times8\times7!}=\frac{1}{132}$

3. $1.2$

Solution:

We have, $\text{P}(\text{A}\cup\text{B})=0.6$ and $\text{P}(\text{A}\cap\text{B})=0.2$

$\therefore\ \text{P}(\text{A}\cup\text{B})=0.6$ and $\text{P}(\text{A}\cap\text{B})=0.2$

$\Rightarrow0.6=\text{P(A)}+\text{P(B)}-0.2$

$\Rightarrow\text{P(A)}+\text{P(B)}=0.8$

$\therefore\ \text{P}(\bar{\text{A}})+\text{P}(\bar{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$

$=2-\big[\text{P(A)}+\text{P(B)}\big]=2-0.8=1.2$

2. $\frac{4}{11}$

Solution:

Total number of alphabets in probability = 11

Number of vowels = 4

$\therefore\ \text{Required probability}=\frac{4}{11}$

3. $\frac{26}{51}$

Solution:

We know that out of 52 playing cards 26 are of red and 26 are of black colour.

$\therefore$ P(both cards of differents colour)

$=\frac{26}{52}\times\frac{26}{51}+\frac{26}{52}\times\frac{26}{51}$

$=2\times\frac{26}{52}\times\frac{26}{51}=\frac{6}{51}$

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

Solution:

Given that, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cap\text{B})$

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

$\Rightarrow\big[\text{P(A)}-\text{P}(\text{A}\cap\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]=0$

But $\text{P(A)}-\text{P}(\text{A}\cap\text{B})\geq0\ ....(\text{i})$

$\big[\because\ \text{P}(\text{A}\cap\text{B})\leq\text{P(A)}\text{ or }\text{P(B)}\big]$

And $\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})\geq0\ ...(\text{ii})$

From eq. (i) and (ii) we get

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

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

Solution:

If two persons sit next to each other, then consider these two person as 1 geoup.

Now, we have to arrange 6 persons.

$\therefore$ Number of arrangement = 2! × 6!

Total number of arrangement of 7 person = 7!

$\therefore\ \text{Required probability}=\frac{2!6!}{7!}=\frac{2}{7}$

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

Solution:

We have digite 0, 2, 3, 5.

Number of divisible by 5 if unit place digit is '0' or '5'

If unit place is '0' then first three places can be filled in 3! ways.

If unit place is '5' then first place can be filled in two ways and second and thried place can be filled in 2! ways.

So, number of numbers ending with digit '5' is 2 × 2! = 4

$\therefore$ Total number of numbers divisible by 5 = 3! + 4 = 110 = n(E)

Also total number of numbers = 3 × 3! = 18

$\therefore\ \text{Required probability}=\frac{10}{18}=\frac{5}{9}$

2. $\frac{187}{190}$

Solution:

Since, the set of three consecutive numbers from 1 to 20 are (1, 2, 3), (2, 3, 4), (3, 4, 5), ....... , (18, 19, 20), i.e., 18

P(numbers are consecutive)

$=\frac{18}{^{20}\text{C}_3}=\frac{18}{\frac{20\times19\times18}{3!}}=\frac{3}{190}$

P(three number are not consecutive)

$=1-\frac{3}{190}=\frac{187}{190}$

3. $\leq0.5$

Solution:

Given, P(A fail) = 0.2

and P(B fail) = 0.3

$\therefore\ \text{P(either A or B fail)}\leq\text{P(A fail)}+\text{P(B fail)}$

$\leq0.2+0.3$

$\leq0.5$

1. $\text{P(A)}\leq\text{P}(\bar{\text{B}})$

Solution:

For mutually exclusive events,

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

$\therefore\ \text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}$ $\big[\because\text{ P}(\text{A}\cap\text{B})=0\big]$

$\Rightarrow\text{P}(\text{A})+\text{P(B)}\leq1$

$\Rightarrow\text{P(A)}+1-\text{P}(\bar{\text{B}})\leq1$ $\big[\text{P(B)}=1-\text{P}(\bar{\text{B}})\big]$

$\Rightarrow\text{P(A)}-\text{P}(\bar{\text{B}})\leq0$

$\Rightarrow\text{P(A)}-\text{P}(\bar{\text{B}})$

2. $\text{P(M)}+\text{P(N)}-\text{P(M}\cap\text{N)}$

Solution:

If M and N are any two events.

$\text{P(M}\cup\text{N)}=\text{P(M)}+\text{P(N)}-\text{P(M}\cap\text{N)}$