Mathematics

Probability Distributions

457 Questions

Probability distributions describe the likelihood of different outcomes in a statistical experiment. Key topics include calculating confidence intervals, understanding Type II errors, and working with the standard normal distribution. These concepts are frequently tested in mathematics and statistics sections of various competitive exams.

Standard normal distributionType II error probabilityConfidence interval interpretationPoisson distributionProbability density function

Probability Distributions Questions

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

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

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

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

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. $\dfrac{1}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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

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

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. $0.592$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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.