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Some more terms in probability - class-XI

Description: some more terms in probability
Number of Questions: 75
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Tags: maths introduction of probability theory probability statistics statistics and probability
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The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

  1. $0$

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

  3. $\displaystyle\frac { 1 }{ 12 } $

  4. $\displaystyle\frac { 1 }{ 36 } $


Correct Option: D
Explanation:

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).

If the letters of the word  $"ATTEMPT"$  are written down at random. The probability that all the  $T's$  come together is

  1. $1/21$

  2. $6/7$

  3. $1/7$

  4. $1/42$


Correct Option: C
Explanation:

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

Number of ways of arranging the words  is  $\dfrac{7!}{3!}$
Probability that all the $T's$  together is $\dfrac{5! }{\dfrac{7!}{3!}}=\dfrac{1}{7}$

The probability of getting number 10 in a throw of a dice is ____.

  1. 0

  2. 1

  3. 0.5

  4. 0.75


Correct Option: A
Explanation:

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

The probability of _____ event is 0.

  1. Sure

  2. Impossible

  3. Exclusive

  4. None of these


Correct Option: B
Explanation:

The probability of an impossible event is 0.

The probability of ____ event is 1.

  1. Sure

  2. Impossible

  3. exclusive

  4. mutually exclusive


Correct Option: A
Explanation:

The probability of a sure event is 1.

A bag contains $4$ red balls, $6$ blue balls and $3$ black balls. A ball is draw at random from the bag. What is the probability that the ball drawn is not blue?

  1. $\displaystyle\frac{6}{13}$

  2. $\displaystyle\frac{3}{13}$

  3. $\displaystyle\frac{7}{13}$

  4. None


Correct Option: C
Explanation:

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

Total no. of ways one ball out of 7,n(S)$=13C _1$
No.of ways of drawing 1 ball,none of then is blue,n(E)$=13- 6=7C _1$
$\therefore  probability=\dfrac{n(E)}{n(S)}=\frac{7}{13}$

The probability of a certain event is 

  1. $0$

  2. $1$

  3. greater than $1$

  4. less than $0$


Correct Option: B
Explanation:
An event which always happens is called a sure event or a certain event. So the probability of a certain event is $1$. 
For example, when we throw a die, then the event "getting a number less than $7$" is a certain event.

If P(E) = 0 then E is a/an

  1. sure event

  2. impossible event

  3. equally likely event

  4. none of these


Correct Option: B
Explanation:

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

If P(E)=0 then the event is called impossible event.
For example -
When a dice is thrown the possible outcomes are 1,2,3,4,5 and 6.
then  the probability is to getting the number 7  in a single throw of a dice is 0 then this is called impossible event.
$P(E)=\frac{0}{6}=0$    

The probability of an impossible event is 

  1. $1$

  2. $0$

  3. less than $0$

  4. greater than $1$


Correct Option: B
Explanation:

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

So, the probability of an impossible event is always zero.

The event which cannot happen is called 

  1. outcome

  2. impossible event

  3. frequency

  4. none of these


Correct Option: B
Explanation:

Event which cannot happen is called impossible event.

Any subset of sample space is called

  1. event

  2. probability

  3. outcome

  4. exprement


Correct Option: A
Explanation:

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

Which one of the following is an impossible event?

  1. Rolling a die to get $4$

  2. Tossing a coin to get tail

  3. Choosing $4$ face cards of spades.

  4. Rolling a die for $7$.


Correct Option: C,D
Explanation:

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)

Rolling a die gives outcomes as $= {1, 2, 3, 4, 5, 6}$.

Choosing a queen from a deck of cards is an example of

  1. compound event

  2. complementary event

  3. simple event

  4. impossible event


Correct Option: A
Explanation:

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 ___________.

  1. simple event

  2. compound event

  3. complementary event

  4. certain event


Correct Option: D
Explanation:

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 _____ is greater than or equal to $0$ and less than or equal to $1$.

  1. space

  2. experiment

  3. sample

  4. event


Correct Option: D
Explanation:

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 _____ connected with the experiment.

  1. space

  2. events

  3. experiment

  4. random


Correct Option: B
Explanation:

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 ____.

  1. compound event

  2. simple event

  3. impossible event

  4. complementary event


Correct Option: B
Explanation:

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

  1. compound

  2. simple

  3. impossible

  4. complementary


Correct Option: A
Explanation:

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 $E = {2, 4, 6}$.

What is called one or more outcomes of an experiment?

  1. Space

  2. Experiment

  3. Sample

  4. Event


Correct Option: D
Explanation:

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

A die is rolled, find the probability that an odd numbers is obtained.

  1. $\dfrac{1}{2}$

  2. $\dfrac{3}{2}$

  3. $\dfrac{7}{2}$

  4. $\dfrac{6}{3}$


Correct Option: A
Explanation:

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

  1. impossible event

  2. sure event

  3. exhaustive event

  4. complementary


Correct Option: A
Explanation:

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 which is called

  1. An impossible event

  2. A sure event

  3. An exhaustive event

  4. A complementary event


Correct Option: B
Explanation:

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 $P(\phi) =$ ?

  1. $0$

  2. $1$

  3. $\phi$

  4. $-1$


Correct Option: A
Explanation:

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

Two cards are drawn from a single deck of $52$ cards one after the other. Find the probability of selecting a king from the first card and queen from the second card.

  1. $\dfrac{1}{26}$

  2. $\dfrac{4}{52}$

  3. $\dfrac{16}{663}$

  4. $\dfrac{4}{663}$


Correct Option: D
Explanation:

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

Probability of selecting a queen from the second card after the first card is drawn out is $\dfrac{4}{51}$. (Since $4$ queens in left over $51$ cards.
Now probability of selecting a king from the first card and queen from the second card is $\dfrac{4}{52}\times \dfrac{4}{51}=\dfrac{4}{663}$.
Hence, option D is correct.

Toss three fair coins simultaneously and record the outcomes. Find the probability of getting atmost one head in the three tosses.

  1. $\dfrac{1}{6}$

  2. $\dfrac{1}{4}$

  3. $\dfrac{1}{2}$

  4. $\dfrac{1}{3}$


Correct Option: C
Explanation:

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

atmost one head in 4 events
the probability of getting atmost one head in the three tosses.$P(E)=\dfrac{4}{8}=\dfrac{1}{2}$

Which one of the following is correct?

  1. An event having no sample point is called an elementary event

  2. An event having one sample point is called an elementary event

  3. An event having two sample point is called an elementary event

  4. An event having many sample point is called an elementary event


Correct Option: B
Explanation:

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

Identify and write the like terms in each of the following groups.
(i) $ a^2, b^2, -2a^2 , c^2 , 4a$ 

  1. $(a^6,2a^2)$

  2. $(a^2,-2a^2)$

  3. $(a^3,2a^2)$

  4. $(a^2,2a^3)$


Correct Option: B
Explanation:

In $a^{2},b^{2},-2a^{2},c^{2},4a.$

 $a^{2}$ and  $-2a^{2}$ are like terms because  $-2a^{2}$ is a factor of $a^{2}$ 
$B$ is correct.

$P\left(\dfrac{B}{ A}\right)$ is defined only when:

  1. $A$ is a sure event

  2. $B$ is a sure event

  3. $A$ is not an impossible event

  4. $B$ is an impossible event


Correct Option: C
Explanation:

$P(\dfrac{B}{A})$ 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 defined only when

  1. $B$ is not a sure event

  2. $B$ is a sure event

  3. $B$ is an impossible event

  4. $B$ is not an impossible event


Correct Option: A
Explanation:

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). 

Similarly, P(A/B') is possible only when B' is sure or B is not sure.

If $P(A) = 1$, then the event $A$ is known as

  1. Symmetric event

  2. Dependent event

  3. Improbable event

  4. Sure event


Correct Option: D
Explanation:

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.

If $P(A) = 0$, then the event $A$

  1. Will never happen

  2. Will always happen

  3. May happen

  4. May not happen


Correct Option: A
Explanation:

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

The probability of a sure event (or certain event) is ____

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: B
Explanation:

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

If the number of desirable events is equal to the total number of possible events, then probability of an event which is certain or sure to happen $ = 1$
Such an event is called certaind event or sure event.

The probability of an event that is certain to happen is ____?

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: A
Explanation:

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.

Tickets numbered from $1$ to $30$ are mixed up and then a ticket is drawn at random. What is the probability that the drawn ticket has a number which is divisible by both $2$ and $6$?

  1. $\dfrac{1}{2}$

  2. $\dfrac{2}{5}$

  3. $\dfrac{8}{15}$

  4. $\dfrac{1}{6}$


Correct Option: D
Explanation:

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


Let $E$ be the event of number divisible by both $2$ and $6$.

$E={6,12,18,24,30}$

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{5}{30}=\dfrac{1}{6}$

$\therefore$   the probability that the drawn ticket has a number which is divisible by both $2$ and $6$ is $\dfrac{1}{6}$

The number of ways in which $6$ men can be arranged in a row, so that three particular men are consecutive, is 

  1. $4! \times 3!$

  2. $4!$

  3. $3! \times 3!$

  4. none of these


Correct Option: A
Explanation:
To arrange $6$ men in a row such three particular men are consecutive $M _1M _2M _3M _4M _5M _6$
three men to be consecutive lets make $3$ men in a Group
$(M _1M _2M _3)$ $M _4M _5M _6$
Total no. of ways of arranging them is $4!\times3!$.

If A and B are such events that $P(A)>0$ and $ P(B)\neq 1$ then $P\left(\dfrac{\bar{A}}{\bar{B}}\right)$ is equal to-

  1. $1-P\left(\dfrac{A}{B}\right)$

  2. $1-P\left(\dfrac{\bar{A}}{B}\right)$

  3. $\dfrac{1-P(A\cup B)}{P(\bar{B})}$

  4. $None$


Correct Option: C

4 normal distinguishable dice are rolled once. The number of possible outcomes in which at least one dice shows up 2?

  1. 216

  2. 648

  3. 625

  4. 671


Correct Option: D
Explanation:
The number of possible outcomes in which atleast one dice shows up $2$ is:
$\begin{array}{l} =(6\times 6\times 6\times 6)-(5\times 5\times \times 5\times 5) \\= 1296-625=671 \end{array}$

Hence, the correct option is $D$

$8$ players compete in a tournament, every one plays everyone else just once. The winner of a game gets $1$, the loser $0$ or each gets $\dfrac{1}{2}$ if the game is drawn. The final result is that every one gets a different score and the player playing placing second gets the same as the total of four bottom players.The total score of all the players is

  1. $28$

  2. $21$

  3. $20$

  4. $22$


Correct Option: A
Explanation:

The players get $1$ for winning 

and $\dfrac 12$ for a draw 
So the sum of points of the players for one match is $1$
Hence the total score for $^8C _2=\dfrac{8\times7}2$ matches is $28$
'

A fair die is thrown 3 times . The chance that sum of three numbers appearing on the die is less than 11 , is equal to -

  1. $\dfrac{1}{2}$

  2. $\dfrac{2}{3}$

  3. $\dfrac{1}{6}$

  4. $\dfrac{5}{8}$


Correct Option: A
Explanation:
From given, we have,

Sum 3:$ (1, 1, 1)$ ==> Contributing only $1$ distinct triplet.

Sum 4: $(1, 1, 2)$ ==> Contributing $3$ distinct triplets

Sum 5: $(1, 2, 2) $and $(1, 1, 3)$ ==> Contributing $6$ distinct triplets

Sum 6: $(1, 1, 4), (1, 2, 3)$ and $(2, 2, 2)$ => Together contributing $10 $distinct triplets

Sum 7: $(1, 1, 5), (1, 2, 4), (1, 3, 3) $and $(2, 2, 3)$ => Together contributing $15$ distinct triplets.

Sum 8: $(1, 1, 6), (1, 2, 5), (1, 3, 4), (2, 3, 3)$ and$ (2, 4, 2) $==> Together contributing $21$ distinct triplets.

Sum 9: $(1, 2, 6), (1, 3, 5), (1, 4, 4), (2, 3, 4), (2, 5, 2)$ and $(3, 3, 3)$ ==> Together contributing $25$ distinct triplets.

Sum 10: $(1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4)$ and $(3, 3, 4) $==> Together contributing $27$ distinct triplets.

Therefore number of favorable cases $= 1+ 3 + 6 + 10 + 15 + 21 + 25 + 27 = 108.$

Therefore, probability $= \dfrac{108}{216} =\dfrac{1}{2}$

The probability that a number selected at random from the numbers $1,2,3.......15$ is a multiple of $4$ is 

  1. $\dfrac{4}{15}$

  2. $\dfrac{2}{15}$

  3. $\dfrac{1}{15}$

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


Correct Option: D
Explanation:

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, to each of which corresponds an envelope, are placed in the envelopes at random. The probability that all the letters are not placed in the right envelopes, is

  1. $\dfrac{1}{6}$

  2. $\dfrac{5}{6}$

  3. $\dfrac{1}{3}$

  4. $\dfrac{2}{3}$


Correct Option: B
Explanation:

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$=1-\dfrac{1}{3!}=\dfrac{5}{6}$

A coin is tossed and a single $6$-sided die is rolled. Find the probability of landing on the tail side of the coin and rolling $4$ on the die.

  1. $\dfrac{1}{12}$

  2. $\dfrac{6}{5}$

  3. $\dfrac{4}{3}$

  4. $\dfrac{3}{4}$


Correct Option: A
Explanation:

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

$P$ (tail and $4$) $=$ $P$(tail) $. P(4)$
$=$$\cfrac{1}{2}\times \cfrac{1}{6}$ $=$ $\cfrac{1}{12}$

The probability of getting number less than or equal to $6$, when a die is thrown once, is

  1. An impossible event

  2. A sure event

  3. An exhaustive event

  4. A complementary event


Correct Option: B
Explanation:

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 $1, 2, 3, 4, 5$.

Two dice are tossed once. The probability of getting an even number at the first die or a total of $8$ is

  1. $\dfrac{1}{36}$

  2. $\dfrac{3}{36}$

  3. $\dfrac{11}{36}$

  4. $\dfrac{20}{36}$


Correct Option: D
Explanation:

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

Calculate the probability that a number selected at random from the set {$2,3,7,12,15,22,72,108$} will be divisible by both $2$ and $3$.

  1. $\cfrac{1}{4}$

  2. $\cfrac{3}{8}$

  3. $\cfrac{3}{5}$

  4. $\cfrac{5}{8}$

  5. $\cfrac{7}{8}$


Correct Option: B
Explanation:

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

In total there are $8$ numbers in the sample set. 
Therefore the required probability is $\dfrac{3}{8}$.

Two similar boxes $B _{i}(i = 1, 2)$ contains $(i + 1)$ red and $(5 - i - 1)$ black balls. One box is chosen at random and two balls are drawn randomly. What is the probability that both the balls are of different colours?

  1. $\dfrac{1}{2}$

  2. $\dfrac{3}{10}$

  3. $\dfrac{2}{5}$

  4. $\dfrac{3}{5}$


Correct Option: D
Explanation:
Clearly, $B _1$ has 2 red and 3 black balls whereas $B _2$ has 3 red and 2 black balls.
The probability of choosing a box randomly is $\dfrac{1}{2}$.
Assume that $B _1$ is chosen first and that red ball is drawn first and black ball second. Then, the probability of this event is $\dfrac{1}{2}\times\dfrac{2}{5}\times\dfrac{3}{4}=\dfrac{3}{20}$
Now, assume that $B _1$ is chosen first and that black ball is drawn first and red ball second. Then, the probability of this event is $\dfrac{1}{2}\times\dfrac{3}{5}\times\dfrac{2}{4}=\dfrac{3}{20}$
Now, assume that $B _2$ is chosen first and that black ball is drawn first and red ball second. Then, the probability of this event is $\dfrac{1}{2}\times\dfrac{2}{5}\times\dfrac{3}{4}=\dfrac{3}{20}$
Now, assume that $B _2$ is chosen first and that red ball is drawn first and black ball second. Then, the probability of this event is $\dfrac{1}{2}\times\dfrac{3}{5}\times\dfrac{2}{4}=\dfrac{3}{20}$
Hence, probability that one box is picked at random and the outcome of draw is 2 balls of different color is the sum of all the above described events=$4\times\dfrac{3}{20}=\dfrac{3}{5}$.
This is the required solution.

Simone and her three friends were deciding how to pick the song they will sing for their school's talent show. They decide to roll a number cube.
The person with the lowest number chooses the song. If her friends rolled a 6, 5, and 2, what is the probability that Simone will get to choose the song?

  1. $\dfrac{1}{6}$

  2. $\dfrac{1}{3}$

  3. $0$

  4. $1$


Correct Option: A
Explanation:

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


In order for Simone to be able to choose the song, she will need to roll a $1$. 

Let $P(A)$ be the probability that Simone chooses the song. 

$P(A) = \dfrac{number:of:favorable:outcomes}{number:of:possible:outcome}$
           $=\dfrac{1}{6}$ (there are $6$ possible outcomes, and $1$ of them is favorable) 

The probability that Simone will choose the song is $\dfrac{1}{6}$.

A box contains $6$ green balls, $4$ blue balls and $5$ yellow balls. A ball is drawn at random. Find the probability of
(a) Getting a yellow ball.
(b) Not getting a green ball.

  1. $\dfrac{1}{5},\dfrac{1}{3}$

  2. $\dfrac{4}{15}, \dfrac{3}{15}$

  3. $\dfrac{1}{3}, \dfrac{3}{5}$

  4. $\dfrac{2}{3}, \dfrac{1}{15}$


Correct Option: C
Explanation:

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


Total number of balls $n(S)=6+4+5=15$

$(a)$ 

Let $A$ be the probability of getting yellow ball.


$n(A)=5$

Thus the probability of getting yellow ball is $P(A)=\dfrac{n(A)}{n(S)}=\dfrac{5}{15}=\dfrac{1}{3}$.

$(b)$

Let $B$ be the probability of not getting green ball. That is, probability of getting blue and yellow balls.

$n(B)=4+5=9$

Thus the probability of getting yellow ball is $P(B)=\dfrac{n(B)}{n(S)}=\dfrac{9}{15}=\dfrac{3}{5}$.

A researcher conducted a survey to determine whether people in a certain town prefer watching sports on television to attending the sporting event. The researcher asked 117 people who visited a local restaurant on a Saturday, and 7 people refused to respond. Which of the following factors makes it least likely that a reliable conclusion can be drawn about the sports-watching preferences of all people in the town?

  1. Sample size

  2. Population size

  3. The number of people who refused to respond

  4. Where the survey was given.


Correct Option: A
Explanation:

Considering the population of a town, the number  $117$  is very much low.

We can't take a survey of  $117$  people and conclude the result for the whole town.
So, the  $sample\ size$  of the survey is very low compared to the population.   $[A]$

Sita and Geta are friends, what is the probability that both will have different birthdays (ignoring a leap year)

  1. $\dfrac { 1 }{ 365 } $

  2. $\dfrac { 1 }{ 364 } $

  3. $\dfrac { 364 }{ 365 } $

  4. None of these


Correct Option: C
Explanation:

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

The probability that an event does not happens in one trial is 0.8.The probability that the event happens atmost once in three trails is 

  1. $0.896$

  2. $0.791$

  3. $0.642$

  4. $0.592$


Correct Option: A

If for two events $A$ and $B, P(A\cap B)\ne P(A) \times P(B)$, then the two events $A$ and $B$ are

  1. Independent

  2. Dependent

  3. Not equally likely

  4. Not exhaustive


Correct Option: B
Explanation:

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

So, $P\left( A\cap B \right) \neq P\left( A \right) .P\left( B \right) $ implies that A and B are independent.

A bag contains four tickets marked with $112, 121, 211, 222$, one ticket is drawn at random from the bag. Let $E _i(i=1, 2, 3)$ denote the event that $i^{th}$ digit on the ticket is $2$ then :

  1. $E _1$ and $E _2$ are independent

  2. $E _2$ and $E _3$ are independent

  3. $E _3$ and $E _1$ are independent

  4. $E _1, E _2, E _2$ are independent


Correct Option: A,B,C
Explanation:

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


$P(E _i \cap E _j) = \dfrac{1}{4} = P(E _i)P(E _j)$

Hence, two events taken together are independent.

$P(E _1 \cap E _2 \cap E _3) = \dfrac{1}{4} \neq P(E _1)P(E _2)P(E _3)$

Therefore, the three events are not independent together.

Hence, options A, B and C are correct.

Two cards are drawn simultaneously from a well shuffled pack of $52$ cards. The expected number of aces is?

  1. $\dfrac{1}{221}$

  2. $\dfrac{3}{131}$

  3. $\dfrac{2}{113}$

  4. $\dfrac{1}{131}$


Correct Option: A
Explanation:

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

Probability of any event $x$ lies

  1. $0 < x < 1$

  2. $0\leq x < 1$

  3. $0\leq x \leq 1$

  4. $1 < x < 2$


Correct Option: C
Explanation:

$P\in \left [ 0,1 \right ]$ 


Option $(C)$ is correct.

Probability of impossible event is

  1. $1$

  2. $0$

  3. $\dfrac {1}{2}$

  4. $-1$


Correct Option: B
Explanation:

(P) Of possible event is $0$

Which one can represent a probability of an event

  1. $\dfrac {7}{4}$

  2. $-1$

  3. $-\dfrac {2}{3}$

  4. $\dfrac {2}{3}$


Correct Option: D
Explanation:

     $P <  1$


 Optoion (D) is correct $2/3$

Probability of sure event is

  1. $1$

  2. $0$

  3. $\dfrac {1}{2}$

  4. $2$


Correct Option: A
Explanation:

Probability of sure event is always $1$

If P(A) = P(B), then

  1. A and B are the same events

  2. A and B must be same events

  3. A and B may be different events

  4. A and B are mutually exclusive events.


Correct Option: C
Explanation:
Given $P(A) = P(B)$
Then $A$ can be different event from $B$ because $A\;\xi\;B$ are Mutually exclusive.
i.e, $A\;\xi\;B$ may be different events.
Hence, the answer is $A$ and $B$ may be different events.

In a single throw of die, what is the probability of getting a number greater than 3 ?

  1. $\dfrac12$

  2. $\dfrac23$

  3. $\dfrac13$

  4. none of these


Correct Option: A

An urn contains 2 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.

  1. $\dfrac57$

  2. $\dfrac{10}{21}$

  3. $\dfrac27$

  4. $\dfrac{11}{21}$


Correct Option: B

If the chance that a vessel arrives safely at a port is $\dfrac 9{10}$ then what is the chance that out of $5$ vessels expected at least $4$ will arrive safely?

  1. $\dfrac {14 \times 9^4}{10^5}$

  2. $\dfrac {15 \times 9^5}{10^4}$

  3. $\dfrac {14 \times 9^3}{10^4}$

  4. $\dfrac {14 \times 9^6}{10^5}$


Correct Option: A

A box contains nine bulbs out of which $4$ are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.

  1. $\dfrac{62}{63}$

  2. $\dfrac{125}{126}$

  3. $\dfrac{1}{63}$

  4. $\dfrac{1}{126}$


Correct Option: D

A pot has $2$ white, $6$ black, $4$ grey and $8$ green balls. If one ball is picked randomly from the pot, what is the probability of it being black or green?

  1. $\dfrac34$

  2. $\dfrac1{10}$

  3. $\dfrac43$

  4. $\dfrac7{10}$


Correct Option: D

$10$ books are placed at random in a shelf. The probability that a pair of books will always be together is 

  1. $\dfrac1{5}$

  2. $\dfrac9{10}$

  3. $\dfrac1{10}$

  4. $\dfrac3{10}$


Correct Option: A

A basket contains $6$ blue, $2$ red, $4$ green and yellow balls. If three balls are picked up at random, what is the probability that none is yellow ?

  1. $\dfrac{3}{455}$

  2. $\dfrac{1}{5}$

  3. $\dfrac{4}{5}$

  4. $\dfrac{44}{91}$


Correct Option: D

Successive trials in binomial distribution are

  1. Dependent

  2. Independent

  3. Equally Likely

  4. Mutually exclusive


Correct Option: B

The probability that A speaks truth is $\dfrac35$ and that of B speaking truth is $\dfrac47$. What is the probability that they agree in stating the same fact?

  1. $\dfrac {18}{35}$

  2. $\dfrac {12}{35}$

  3. $\dfrac {17}{35}$

  4. $\dfrac {19}{35}$


Correct Option: A

If the occurrence of one event means that another cannot happen, then the events are

  1. Bayesian

  2. Empirical

  3. Independent

  4. Mutually Exclusive


Correct Option: D

The probability of success of three students X,Y and Z in the one examination are $\dfrac15, \dfrac14$ and $\dfrac13$ respectively. Find the probability of success of at least two.

  1. $\dfrac16$

  2. $\dfrac25$

  3. $\dfrac34$

  4. $\dfrac35$


Correct Option: A

For the special rule of multiplication of probability, the events must be

  1. Empirical

  2. Bayesian

  3. Independent

  4. none of these


Correct Option: C

In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11 ?

  1. $\displaystyle \frac{7}{12}$

  2. $\displaystyle \frac{5}{36}$

  3. $\displaystyle \frac{1}{6}$

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


Correct Option: B
Explanation:
Let A denotes the event of getting a total of 10 and B denotes the event of getting a total of 11.
Total outcomes = $6\times 6 $
A= {(5,5),(6,4),(4,6)}
n(A) = $3$
$\therefore $ p(A) = $\dfrac {3}{36}  $
B={(6,5),(5,6)}
n(B) = 2
$\therefore $ p(B)= $\dfrac {2}{36}  $
$\therefore $ p(A or B) = p(A) + p(B) 
                   = $\dfrac {3}{36} + \dfrac {2}{36} $
                   = $\dfrac {5}{36} $
Option B is correct.

How many times must a man toss a fair coin, so that the probability of having at least one head is more than $80 \%?$

  1. $3$

  2. $>3$

  3. $<3$

  4. none of these


Correct Option: B
Explanation:
In any fair coin toss, P (getting a head) = P (getting a tail) i.e., p=q=$\dfrac 12$
We need to find n such that the probability of getting at least one head is more than $80\%$
$P(X≥1)=1−P(X<1)>80\%$
$\implies 1−P(X=0)>\dfrac 8{10}\\\implies P(X=0)<1−\dfrac 8{10}\\\implies P(X=0)<\dfrac 2{10} or P(X=0)<\dfrac 15$
For a bionomial distribution, $P(X=0)=^nC _0\left(\dfrac 12\right)^0\left(\dfrac 12\right)^{n−0}=\left(\dfrac 12\right)^n$
$\implies \left(\dfrac 12\right)^n<\dfrac 1{5}\\\implies 2^n>5$
Since $2^1=2,2^2=4, 2^3=8,2^4=16$, the minimum value for n that satisfies the inequality is $n=3$, i.e, the coin should be tossed $3$ or more times.

Calculate the probability that a spinner, having the numbers one through five evenly spaced, will land on an odd number exactly once if the spinner is used three times.

  1. $\dfrac {12}{125}$

  2. $\dfrac {18}{125}$

  3. $\dfrac {27}{125}$

  4. $\dfrac {36}{125}$

  5. $\dfrac {54}{125}$


Correct Option: D
Explanation:

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 $\dfrac {36}{125}$.

A bag contains $10$ balls, each labelled with a different integer from $1$ to $10$, inclusive. If $2$ balls are drawn simultaneously from the bag at random, calculate the probability that the sum of the integers on the balls drawn will be greater than $6$.

  1. $0.41$

  2. $0.43$

  3. $0.60$

  4. $0.76$

  5. $0.87$


Correct Option: E
Explanation:

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 $1-\dfrac {6}{45} = \dfrac {39}{45} = 0.87$

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