Addition theorems of probability - class-XI

Here, $\displaystyle A=\left{ 4,5,6\right} $
and  $\displaystyle B=\left{ 1,2,3,4,\right} $
We have $\displaystyle A\cup B=\left{1,2,3,4,5,6\right} =S$
Where S is the sample space of the experiment of throwing a die.
$\displaystyle P(S) =1$ since it is a sure event.Hence $\displaystyle P(A\cup B)=1$

Total number of bags = $12 + 5 + 2 + 1 = 20$
Bags which weigh more than $10 kg = 12$
Thus Probability (bag weighs more than 10 kg) = $\dfrac{12}{20} = 0.6$

We know that the sum of all the elementary events = 1
$\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7} + P _3 = 1$ 

$\Rightarrow \quad P _3 = 1 - \left(\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7}\right) = \displaystyle\frac{4}{7}$ 

No. of Kings of heart $=$ 1, No. of Queen of diamonds $=$1
Probability of king of hearts or queen of diamonds $ = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$

One card is both a 5 and a Club.
$\therefore$ Club or numbered 5 are not mutually exclusive.
P(club or numbered 5) $\neq$ 0.4 + 0.3

Total number of students $= 400$
No. of science students $= 187$
No. of commerce students $ = 125$
Thus, Probabilty of the student being either science or commerce = $\dfrac{187}{400} + \dfrac{125}{400}$
= $\dfrac{312}{400}$
= $\dfrac{39}{50}$

Let A=the integer is divisible by 6 
$A={6,12,18,....198}$
$198=6+(n-1)6$
$\Rightarrow n =33$
So, $P(A)=\dfrac{33}{200}$

B=the integer is divisible by 8
$B={8,16,24,....200}$
$200=8+(n-1)8$
$\Rightarrow n=25$
So, $P(B)=\dfrac{25}{200}$

Both are divisible by 6 and 8 both 
$A\cap B=24,48,.....192$
$192=24+(n-1)24$
$\Rightarrow n=8$
So, $P(A\cap B)=\dfrac{8}{200}$

$\displaystyle \therefore P\left ( A\cup B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( AB \right )$
$\displaystyle=\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{1}{4}.$

Simply speaking, any number on the dice is either less than $5$ or more than $3$ (since the range covers all the numbers from $1$ to $6$). Hence, required probability = $1$.

Alternate method:
$P(A\cup  B)=P(A)+P(B)-P(A\cap B)=\dfrac { 3 }{ 6 } +\dfrac { 4 }{ 6 } -\dfrac { 1 }{ 6 } =1$

$n(A\cup B)={1,2,3,4,5,6}\therefore  P(A\cup B)=1$

$\displaystyle P(A)=\frac { 1 }{ 52 } +\frac { 1 }{ 52 } =\frac { 2 }{ 52 } =\frac { 1 }{ 26 } $

Total Number of Balls $= 5+8+10 = 23$ balls

Probability of getting a red ball  $= \dfrac{5}{23} $

Probability of getting a green ball  $= \dfrac{8}{23} $

Probability of getting a pink ball  $= \dfrac{10}{23} $

Then, probability of getting a red  or a green ball  $= \dfrac{5+8}{23} = \dfrac{13}{23}$

There would be three cases.

$x+\dfrac {100}x>50\\implies x^2 +100>50x\\implies x^2-50x+100+(25)^2-(25)^2>0\\implies x^2+(25)^2-50x-525>0\\implies (x-25)^2>525\\implies x-25>\pm\sqrt{525}\\implies x>25+22.91, x<25-22.91\\implies x>47.91, x<2.09\\implies x\ge 48, x\le 2$

Both will speak the same thing means both can say truth or both can say false

Let the outcomes of two dice be $x,y$.

The number of possible combinations with two dice is $6 \times 6=36$.

$A\quad and\quad B\quad are\quad mutually\quad exclusive\quad events\quad than\quad P(A\cap B)=0\ P(A)=\frac { 3 }{ 5 } \quad and\quad P(B)=\frac { 1 }{ 5 } \ P(A\cup B)=P(A)+P(B)-P(A\cap B)\ \qquad \qquad =\frac { 3 }{ 5 } +\frac { 1 }{ 5 } -0=\frac { 4 }{ 5 } $

Given:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\ \Rightarrow P(A\cup B)=1-P(\overset { \ _  }{ A } )+P(B)-P(A\cap B)\ \Rightarrow \dfrac { 3 }{ 4 } =1-\dfrac { 2 }{ 3 } +P(B)-\dfrac { 1 }{ 4 } \ \Rightarrow P(B)=\dfrac { 2 }{ 3 } $
Hence, option B is correct.

Given: first $30$ natural numbers

Given that $P\left( A \right) =\cfrac { 1 }{ 8 } ,\quad P\left( B \right) =\cfrac { 5 }{ 8 } $

The women can be seated in $^{2n}C _n$ ways. Let A denote the event that pairs occupy the seats (1, 2), (3, 4), ...., (2n-1, 2n)
and B denote the event that pairs occupy the seats (2, 3), (4, 5), (6, 7), ...., (2n-2, 2n-1), (2n-1).
The number of cases favourable to A (B) is $2^n$. (For each man in the pair there are two choices.)
However, there are just two cases common to A and B. One is the case $\left {(M, W), (M, W), ...(M, W)\right }$ of A(B) and $\left {(W, M), (W, M), ..., (W, M)\right }$ of B(A).
Therefore, $P(A\cup B)=P(A)+P(B)-P(A\cup B)$
$\displaystyle =\frac {2^n+2^n-2}{^{2n}C _n}=\frac {2^{n+1}-2}{^{2n}C _n}$.

If we rolled dice around 500 times then it can be considered as its expected outcome.
Hence,
Probability of obtaining between 8 and 12 as sum = 53 + 46 +28/500= 0.254