Tag: maths

Questions Related to maths

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

maths introduction of probability theory types of events some more terms in probability events and its algebra

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

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

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

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

Show answer
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 $3$.

maths introduction of probability theory types of events some more terms in probability events and its algebra

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

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

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

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

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

Show answer
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?

maths introduction of probability theory types of events some more terms in probability events and its algebra

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

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

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

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

Show answer
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?

maths introduction of probability theory types of events some more terms in probability events and its algebra

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

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

  3. $0$

  4. $1$

Show answer
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.

maths introduction of probability theory types of events some more terms in probability events and its algebra

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

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

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

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

Show answer
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?

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. Sample size

  2. Population size

  3. The number of people who refused to respond

  4. Where the survey was given.

Show answer
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)

maths introduction of probability theory types of events some more terms in probability events and its algebra

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

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

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

  4. None of these

Show answer
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 

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $0.896$

  2. $0.791$

  3. $0.642$

  4. $0.592$

Show answer
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

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. Independent

  2. Dependent

  3. Not equally likely

  4. Not exhaustive

Show answer
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 :

maths introduction of probability theory types of events some more terms in probability events and its algebra

  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

Show answer
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.