Mathematics

Probability

337 Questions

Probability measures the likelihood of an event occurring, such as rolling a specific number on a die or drawing a colored ball. Questions cover simple events, mutually exclusive outcomes, and dice or coin combinations. This topic is consistently asked in mathematics and reasoning sections of competitive exams.

dice probabilitycoin toss eventsdrawing balls probabilitiesplaying card problemsmutually exclusive events

Probability Questions

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

A bag contains  $4$ red,  $3$ black, and  $2$ white balls. If  $2$  balls are selected at random, the probability of selecting atleast one white ball is

  1. $\dfrac { 7 } { 12 }$
  2. $\dfrac { 5 } { 12 }$
  3. $\dfrac { 1 } { 3 }$
  4. $\dfrac { 1 } { 4 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Bag contains $4$ red, $3$ black and, $2$ white balls
Two balls are selected at random
The total no. of ways of doing that is ${ 10 } _{ { C } _{ 2 } }=\dfrac { 10! }{ 8!\times 2! } =\dfrac { 10\times 9\times 8! }{ 8!\times 2! } =\dfrac { 10\times 9 }{ 2 } =45$
Now in the selection we need to ensure that at least on white ball is selected.
Case $1$ :  $1$ white ball $+$ $1$ ball of any other color
This can be done in ${ 2 } _{ { C } _{ 1 } }\times { 7 } _{ { C } _{ 1 } }=2\times 7=14$ ways
Case $2$ :  $2$ white ball
This can be done in ${ 2 } _{ { C } _{ 2 } }=1$ way
$\therefore$   probability of selecting atleast one white ball is $=\dfrac { 14+1 }{ 45 } =\dfrac { 15 }{ 45 } =\dfrac { 1 }{ 3 } $
Answer : Option C.
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

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. $\dfrac{11}{21}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

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

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. $\dfrac7{10}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

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

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. $\displaystyle \frac{1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

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. $\dfrac {54}{125}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\dfrac {36}{125}$.

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

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. $0.87$
Reveal answer Fill a bubble to check yourself
E Correct answer
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 $1-\dfrac {6}{45} = \dfrac {39}{45} = 0.87$

Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

Six coins are tossed $6400$ times. The probability of getting $6$ heads $x$ times using poison distribution is

  1. $6400{e^{ - x}}$
  2. $\frac{{6400{e^{ - x}}}}{{x!}}$
  3. $\frac{{{e^{ - 100}}{{100}^x}}}{{x!}}$
  4. ${e^{ - 100}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Therefore, the probability of getting $6$ heads with $6$ coins $ = {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{64}} = P\left( { > {a _y}} \right)$
Then, 
$\eta P = 6400 \times \frac{1}{{64}} = 100 = m\left( {{a _y}} \right)$
So, by poison's law, $P\left( {x = n} \right) = \frac{{{e^{ - m}}{m^x}}}{{x!}}$
$ = \frac{{{e^{ - 100}}{{100}^x}}}{{x!}}$

Multiple choice

A fair coin is tossed twice. What is the probability of getting two heads?

  1. 1/2

  2. 1/4

  3. 1/8

  4. 1/16
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the coin is fair, the probability of getting heads on each toss is 1/2. The probability of getting two heads is the product of these probabilities, which is (1/2) * (1/2) = 1/4.

Multiple choice

A bag contains 5 red balls, 3 blue balls, and 2 green balls. What is the probability of randomly selecting a blue ball?

  1. 5/10

  2. 3/10

  3. 2/10

  4. 1/10
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The probability of selecting a blue ball is the number of blue balls divided by the total number of balls. So, the probability is 3 / (5 + 3 + 2) = 3/10.

Multiple choice

A random variable X has a probability distribution given by P(X = x) = k * x^2, where k is a constant and x = 1, 2, 3. Find the value of k.

  1. 1/6

  2. 1/3

  3. 1/2

  4. 2/3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of probabilities for all possible values of X must be equal to 1. So, we have: P(X = 1) + P(X = 2) + P(X = 3) = k * (1^2) + k * (2^2) + k * (3^2) = k * (1 + 4 + 9) = k * 14 = 1. Therefore, k = 1/14.

Multiple choice

Two events A and B are independent. If P(A) = 0.4 and P(B) = 0.6, what is the probability of both A and B occurring?

  1. 0.24

  2. 0.16

  3. 0.12

  4. 0.08
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since A and B are independent, the probability of both occurring is the product of their individual probabilities. So, P(A and B) = P(A) * P(B) = 0.4 * 0.6 = 0.24.