Probability Terms and Calculations - Class XI

When two dice are rolled, the number of outcomes is $36$.
The only even prime number is $2$.
Let $E$ be the event of getting an even prime number on each die.
$\therefore E=\left{ \left( 2,2 \right)  \right} $
$\Rightarrow P\left( E \right) =\displaystyle\frac { 1 }{ 36 } $
Therefore, the correct answer is (D).

Number of ways of arranging the words keeping all $T's$ together is  $5!$

Since the outcome of a throw of dice can never be 10, the probability is 0.

The probability of an impossible event is 0.

The probability of a sure event is 1.

Total no. of balls$=4+6+3=13$

$P(E)=\frac{number  of   outcomes  favorable}{Total   numbers  of  possible  outcomes}$

An event that has no chance of occurring is called an impossible event. 

Event which cannot happen is called impossible event.

The set of all the possible outcomes is called the sample space of the experiment and is usually denoted by S. Any subset E of the sample space S is called an event. Here are some examples. Example 1 Tossing a coin

Choosing $4$ face cards in spades as well as rolling a die for $7$ are both impossible events.
The number of face cards in spades $= 3$ (king, queen, jack)

Choosing a queen from a deck of cards is an example of compound event.
Because the total number of card $= 52$
Choosing a queen card $= 4$ (spade, diamond, heart, club)
So, more than one element of the sample space in the set representing an event, then this event is called a compound event.

The probability of an event which is sure to occur at every performance of an experiment is called a certain event.
Example: Head or Tail is a certain event connected with tossing a coin.

The probability of an event is greater than or equal to $0$ and less than or equal to $1$.

The outcomes of a random experiment are called events connected with the experiment.
Example: $1, 2, 3, 4, 5, 6$ are the outcomes of the random experiment of rolling a dice and hence are events connected with it.

When the dice are thrown, the event $E = {4}$ then this event is called simple event.
If there be only one element of the sample space in the set representing an event, then this event is called a simple or elementary event.

The sample space in the set representing an event more than one element is called compound event.
Example: if we throw a dice, having $S = {1, 2, 3, 4, 5, 6}$, the event of a even number being shown is given by $E = {2, 4, 6}$.

Event is called one or more outcomes of an experiment.
Example: rolling a dice, we get a possible outcomes as ${1, 2, 3, 4, 5, 6}$.

Sample space for a die $= {1, 2, 3, 4, 5, 6} = 6$
Odd numbers $= 1, 3, 5 = 3$
So, $P$ (Odd number) $=$ $\dfrac{3}{6}= \dfrac{1}{2}$

An event which will not occur on any account is called an impossible event.
Example: getting $10$ in rolling a die once is an impossible event.

While doing any experiment, there will be a possible outcome is called sure event.
Example: Getting $3$ of $1, 2, 3, 4, 5, 6$ in rolling die is a sure event.

If $\phi$ represents an impossible event, then $P(\phi)= 0$
Because an impossible event is an event that will never occur in an experiment.

Probability of selecting a king in the first card is $\dfrac{4}{52}$. (Since $4$ kings in $52$ cards).

Toss three fair coins simultaneously and record the outcomes.The sample space is HHH, HHT, HTH, HTT, THH, THT, TTH and TTT N=8

An elementary event is an event which contains only a single element in the sample space. So, it will have only $1$ sample point.
Hence, option B is true

$P(\dfrac{B}{A})$ is the conditional probability of $B$ given $A$ or  it is the the probability of $B$ under the condition $A$, which is only possible if event $A$ occurs (i.e., $A$ is a possible event). 

P(A/B) is the conditional probability of A given B or it is the probability of A under the condition B, which is only possible if event B occurs (i.e., B is a possible event). 

The probability is the possibility of an event happening when the probability of an event is 1, it means the event will occur for sure.

The probability is the possibility of an event happening when the probability of an event is zero, it means the event will never occur.

Probability of event in $E$, $ P\left( E \right) =\dfrac { \text {number of events in E} }{ \text {Tota number of possible events} } $

The event that is sure to happen is called a certain event and probability of such an event is $1$ as this event is bound to happen.

Here, $S={1,2,3,4,5,6,.....29,30}$

Using the definition of conditional probability, P(A_bar/B_bar) = P(A_bar intersection B_bar) / P(B_bar). By De Morgan's law, A_bar intersection B_bar = (A union B)_bar, so the numerator is 1 - P(A union B). Thus, the expression is (1 - P(A union B)) / P(B_bar).

The players get $1$ for winning 

From Number$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15$

From 1 to 15, Multiples of 4 are 4, 8, 12 only

So Probability$ = \dfrac {Count \ of \ No. \ which \ are \ multiple \ of \ 4}{Total \ Number \ given}$

$ Prob. = \dfrac{3}{15} = \dfrac{1}{5}$

Option D is correct

Three letters can be placed in 3 envelopes in $3!$ ways, whereas there is only one way of placing them in their right envelopes.
So, probability that all the letters are placed in the right envelopes$=\dfrac{1}{3!}$
Hence, required probability$=1-\dfrac{1}{3!}=\dfrac{5}{6}$

$P$ (tail) $=$ $\dfrac{1}{2}$ and $P(4) =$ $\dfrac{1}{6}$

The probability of getting number less than $6$, when a die is thrown once, is a sure event.
Because, once a die is thrown, sample space $= {1, 2, 3, 4, 5, 6}$
There is a possible event for getting number less than $6$ as outcomes can be $1, 2, 3, 4, 5$.

$A=\text{getting even no on Ist dice}$
$B=\text{getting sum 8}$
So, $n(A)=18$
So, $P(A\cup B)o=\dfrac{18+5-3}{36}=\dfrac{20}{36}$

In the given set the number divisible by both $2$ and $3$ i.e numbers divisible by $6$ are ${12,72,108}$. 

The possible outcomes of rolling a number cube are $1, 2, 3, 4, 5, 6 $. 

A box contains $6$ green balls, $4$ blue balls, $5$ yellow balls.

$P=\dfrac { 365\times 364 }{ 365\times 365 } =\dfrac { 364 }{ 365 } $

The probability of failure in one trial is q = 0.8, so success p = 0.2. For n=3 trials, the probability of at most one success is P(X=0) + P(X=1). P(X=0) = (0.8)^3 = 0.512. P(X=1) = 3 * (0.2)^1 * (0.8)^2 = 3 * 0.2 * 0.64 = 0.384. Summing these gives 0.512 + 0.384 = 0.896.

For independent.events $P\left( A\cap B \right) =P\left( A \right) .P\left( B \right) $

$P(E _1) = P(E _2) = P(E _3) =\dfrac{1}{2}$

$\begin{matrix} Two, , cards, , are, , drawn, , from, , a, , well, , shuffered, , pack, , of, , 52, , card. \ Then,, , number, , access, , is, ,  \ \Rightarrow number, , of, , sample, , space=52/2 \ \Rightarrow There, , are, , four, , ace=\dfrac { { 4/2 } }{ { 52/2 } } =\frac { { 4\times 3 } }{ { 52\times 51 } }  \ =\dfrac { 1 }{ { 13\times 17 } } , , , , , Ans. \  \end{matrix}$

$P\in \left [ 0,1 \right ]$ 

(P) Of possible event is $0$

     $P <  1$

Probability of sure event is always $1$

A standard die has 6 faces: {1, 2, 3, 4, 5, 6}. Numbers greater than 3 are {4, 5, 6}. There are 3 favorable outcomes out of 6 total. Probability = 3/6 = 1/2.

Total balls = 2 red + 3 green + 2 blue = 7. Total ways to draw 2 balls = 7C2 = 21. Favorable outcomes (no blue) = drawing from 5 non-blue balls = 5C2 = 10. Probability = 10/21.

This is a binomial distribution problem where n=5, p=0.9, q=0.1. P(X>=4) = P(X=4) + P(X=5). P(X=4) = 5C4 * (0.9)^4 * (0.1)^1 = 5 * 0.9^4 * 0.1 = 0.5 * 9^4 / 10^4 = 5 * 9^4 / 10^5. P(X=5) = 5C5 * (0.9)^5 = 9^5 / 10^5. Sum = (5 * 9^4 + 9 * 9^4) / 10^5 = 14 * 9^4 / 10^5.

The total number of ways to choose 4 bulbs from 9 is 9C4 = 126. The number of ways to choose 4 defective bulbs from the 4 available is 4C4 = 1. Thus, the probability is 1/126.

Total balls = 2 + 6 + 4 + 8 = 20. The number of black or green balls is 6 + 8 = 14. The probability is 14/20, which simplifies to 7/10.

A defining characteristic of a binomial distribution is that each trial is independent of the others.

They agree if both tell the truth or both lie. P(Both Truth) = (3/5)(4/7) = 12/35. P(Both Lie) = (2/5)(3/7) = 6/35. Total probability = 12/35 + 6/35 = 18/35.

Events that cannot occur simultaneously are defined as mutually exclusive.

Probabilities of success: P(X)=1/5, P(Y)=1/4, P(Z)=1/3. P(Failures): P(X')=4/5, P(Y')=3/4, P(Z')=2/3. At least two success = P(exactly 2) + P(exactly 3). P(exactly 3) = (1/5)(1/4)(1/3) = 1/60. P(exactly 2) = (1/5)(1/4)(2/3) + (1/5)(3/4)(1/3) + (4/5)(1/4)(1/3) = 2/60 + 3/60 + 4/60 = 9/60. Total = 10/60 = 1/6.

The multiplication rule for probabilities, P(A and B) = P(A) * P(B), applies specifically when events are independent.

Total possible combinations when spinner used three times is $5 \times 5 \times 5 = 125$.
Out of three times, inexactly one number is odd implies the other two numbers are even.
The possible combinations such that exactly one number is odd is $3 \times 2 \times 2 + 2 \times 3 \times 2 + 2 \times 2 \times 3 = 36$.
The probability is $\dfrac {36}{125}$.

Out of $10$ balls , $2$ balls can be selected in ${ ^{ 10 }{ C } } _{ 2 } = 45$
Number of ways of selecting $2$ balls such that sum is less than or equal to $6$ is $6$
Probability that the sum of integers on the balls drawn will be greater than $6$ is $1-\dfrac {6}{45} = \dfrac {39}{45} = 0.87$