Mathematics

Probability

303 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 maths probability - iii theorem of total probability bayes theorem probability and probability distribution

There are 50 marbles of 3 colors: blue yellow and black The probability of picking up a blue marble is 3/10 and that of picking up a yellow marble is 1/2 The probability of picking up a black ball is 

  1. 1/5

  2. 1/10

  3. 1/4

  4. 4/5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

P(blue)=$\displaystyle \frac{3}{10}=\frac{15}{50},$i.e., there are 15 blue marbles
P(yellow)=$\displaystyle \frac{1}{2}=\frac{25}{50},$i.e., there are 25 yellow marbles
$\displaystyle \therefore $ Number of black marbles=10
$\displaystyle \therefore $ P(black)=$\displaystyle \frac{10}{50}=\frac{1}{5}$

Multiple choice maths probability - iii theorem of total probability bayes theorem probability and probability distribution

I have three bags that each contain $100$ marbles- Bag $1$ has $75$ red and $25$ blue marbles, Bag $2$ has $60$ red and $40$ blue marbles, Bag $3$ has $45$ red and $55$ blue marbles. I choose one of the bags at random and then pick a marble from the chosen bag, also at random. What is the probability that the chosen marble is red?

  1. $0.60$
  2. $0.40$
  3. $0.50$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that bag 1 contains $75$ red and $25$ blue marbles

Given that bag 2 contains $60$ red and $40$ blue marbles
Given that bag 3 contains $45$ red abd $55$ blue marbles
Now the probability of choosing red ball from bag 1 is $ \dfrac{1}{3} \times \dfrac{75}{100}$
Now the probability of choosing red ball from bag 2 is $ \dfrac{1}{3} \times \dfrac{60}{100}$
Now the probability of choosing red ball from bag 3 is $ \dfrac{1}{3} \times \dfrac{45}{100}$
Now the probability of choosing red ball is $ \dfrac{1}{3}\left(\dfrac{75+60+45}{100}\right)=0.6$ 
Hence, option $A$ is correct.

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then $\displaystyle P(A \cup B)$ is

  1. $\displaystyle \frac{2}{5}$
  2. $\displaystyle \frac{3}{5}$

  3. 0

  4. 1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $\displaystyle A=\left{ 4,5,6\right} $
and  $\displaystyle B=\left{ 1,2,3,4,\right} $
We have $\displaystyle A\cup B=\left{1,2,3,4,5,6\right} =S$
Where S is the sample space of the experiment of throwing a die.
$\displaystyle P(S) =1$ since it is a sure event.Hence $\displaystyle P(A\cup B)=1$

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

A die is thrown. Let $A$ be the event that the number obtained is greater than $3$. Let $B$ be the event that the number obtained is less than $5$. Then $(A\cup B)$ is

  1. $\cfrac{2}{5}$
  2. $\cfrac{3}{5}$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$A \equiv \left\{4, 5, 6\right\}$
$B \equiv \left\{1, 2, 3, 4\right\}$
$\therefore \ A\cup B\equiv \left\{1, 2, 3, 4, 5, 6\right\}$
$\therefore \ P(A\cup B)=\dfrac {6}{6}=1$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

A die is thrown. Let $A$ be the event that the number obtained is greater than $3$. Let $B$ be the event that the number obtained is less than $5$. Then $P(A \cup B)$

  1. $3/5$
  2. $0$
  3. $1$
  4. $2/5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Simply speaking, any number on the dice is either less than $5$ or more than $3$ (since the range covers all the numbers from $1$ to $6$). Hence, required probability = $1$.

Alternate method:
$P(A\cup  B)=P(A)+P(B)-P(A\cap B)=\dfrac { 3 }{ 6 } +\dfrac { 4 }{ 6 } -\dfrac { 1 }{ 6 } =1$

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

A box contains 5 red balls, 8 green balls and 10 pink balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green?

  1. $\dfrac{13}{23}$
  2. $\dfrac{10}{23}$
  3. $\dfrac{11}{23}$
  4. $\dfrac{13}{529}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total Number of Balls $= 5+8+10 = 23$ balls

Probability of getting a red ball  $= \dfrac{5}{23} $

Probability of getting a green ball  $= \dfrac{8}{23} $

Probability of getting a pink ball  $= \dfrac{10}{23} $

Then, probability of getting a red  or a green ball  $= \dfrac{5+8}{23} = \dfrac{13}{23}$

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

In a single throw of two dice, the probability of obtaining a total of $7$ or $9,$ is:

  1. $\dfrac{4}{9}$
  2. $\dfrac{1}{3}$
  3. $\dfrac{7}{18}$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the outcomes of two dice be $x,y$.

For the sum to be $7$ , $x+y=7$. There are $6$ possibilities.
For the sum to be $9$ , $x+y=9$. There are $4$ possibilities.
Total number of possibilities for $7$ or $9$ is $10$.
The probability of obtaining total $7$ or $9$ is $\cfrac{10}{36}=\cfrac{5}{18}$.
Therefore option $D$ is correct.

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

The chance of throwing a total of $3$ or $5$ or $11$ with two dice is:

  1. $\dfrac{5}{36}$
  2. $\dfrac{1}{9}$
  3. $\dfrac{2}{9}$
  4. $\dfrac{19}{36}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The number of possible combinations with two dice is $6 \times 6=36$.

The number of ways of getting $3$ as sum is $2$.
The number of ways of getting $5$ as sum is $4$.
The number of ways of getting $11$ as sum is $2$.
The number of ways of getting $3$ or $5$ or $11$ as sum is $8$.
The probability is $\cfrac{8}{36}=\cfrac{2}{9}$.
Therefore option $C$ is correct.

Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

Two dice each numbered from $1$ to $6$ are thrown together. Let $A$ and $B$ be two events given by
$A:$ even number on the first die
$B:$  number on the second die is greater than $4$

What is $P(A\cup B)$ equal to?

  1. $1/2$
  2. $1/4$
  3. $2/3$
  4. $1/6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given:

Two dice are thrown, hence the total number of all possible ways, n(S) = $6\times 6=36$

A:  even number on the first die
B: the number on the second die is greater than 4

To find:
$P(A\cup B) $
favourable ways of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Hence n(A)=18
$\therefore, P(A)=\dfrac {n(A)}{n(S)}=\dfrac {18}{36}=\dfrac 12$
favourable ways of event B = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 5), (5, 6), (6, 5), (6, 6)}
Hence n(B) = 12
$\therefore, P(B)=\dfrac {n(B)}{n(S)}=\dfrac {12}{36}=\dfrac 13$
Hence, $P(\cup B)=P(A)+P(B)-P(A)(P(B)=\dfrac 12+\dfrac 13-\dfrac 12\times \dfrac 13=\dfrac {3+2-1}6=\dfrac 46=\dfrac 23$