Tag: some more terms in probability

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

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. $\dfrac{2}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$=1-\dfrac{1}{3!}=\dfrac{5}{6}$

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

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. $\dfrac{3}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}$ $=$ $\cfrac{1}{12}$

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $1, 2, 3, 4, 5$.

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

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. $\dfrac{20}{36}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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. $\cfrac{7}{8}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\dfrac{3}{8}$.

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

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. $\dfrac{3}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

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. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac{1}{6}$.

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

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. $\dfrac{2}{3}, \dfrac{1}{15}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $P(B)=\dfrac{n(B)}{n(S)}=\dfrac{9}{15}=\dfrac{3}{5}$.

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

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.
Reveal answer Fill a bubble to check yourself
A Correct answer
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.   $[A]$