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

A bag contains $4$ red balls, $6$ blue balls and $3$ black balls. A ball is draw at random from the bag. What is the probability that the ball drawn is not blue?

  1. $\displaystyle\frac{6}{13}$
  2. $\displaystyle\frac{3}{13}$
  3. $\displaystyle\frac{7}{13}$

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total no. of balls$=4+6+3=13$

Total no. of ways one ball out of 7,n(S)$=13C _1$
No.of ways of drawing 1 ball,none of then is blue,n(E)$=13- 6=7C _1$
$\therefore  probability=\dfrac{n(E)}{n(S)}=\frac{7}{13}$

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

Which one of the following is an impossible event?

  1. Rolling a die to get $4$

  2. Tossing a coin to get tail
  3. Choosing $4$ face cards of spades.
  4. Rolling a die for $7$.
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Choosing $4$ face cards in spades as well as rolling a die for $7$ are both impossible events.
The number of face cards in spades $= 3$ (king, queen, jack)

Rolling a die gives outcomes as $= {1, 2, 3, 4, 5, 6}$.

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

When the dice are thrown, the event $E = {4}$, then this event is called ____.

  1. compound event

  2. simple event

  3. impossible event

  4. complementary event
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the dice are thrown, the event $E = {4}$ then this event is called simple event.
If there be only one element of the sample space in the set representing an event, then this event is called a simple or elementary event.

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

Toss three fair coins simultaneously and record the outcomes. Find the probability of getting atmost one head in the three tosses.

  1. $\dfrac{1}{6}$
  2. $\dfrac{1}{4}$
  3. $\dfrac{1}{2}$
  4. $\dfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Toss three fair coins simultaneously and record the outcomes.The sample space is HHH, HHT, HTH, HTT, THH, THT, TTH and TTT N=8

atmost one head in 4 events
the probability of getting atmost one head in the three tosses.$P(E)=\dfrac{4}{8}=\dfrac{1}{2}$

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

4 normal distinguishable dice are rolled once. The number of possible outcomes in which at least one dice shows up 2?

  1. 216

  2. 648

  3. 625

  4. 671
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The number of possible outcomes in which atleast one dice shows up $2$ is:
$\begin{array}{l} =(6\times 6\times 6\times 6)-(5\times 5\times \times 5\times 5) \\= 1296-625=671 \end{array}$

Hence, the correct option is $D$
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

A fair die is thrown 3 times . The chance that sum of three numbers appearing on the die is less than 11 , is equal to -

  1. $\dfrac{1}{2}$
  2. $\dfrac{2}{3}$
  3. $\dfrac{1}{6}$
  4. $\dfrac{5}{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
From given, we have,

Sum 3:$ (1, 1, 1)$ ==> Contributing only $1$ distinct triplet.

Sum 4: $(1, 1, 2)$ ==> Contributing $3$ distinct triplets

Sum 5: $(1, 2, 2) $and $(1, 1, 3)$ ==> Contributing $6$ distinct triplets

Sum 6: $(1, 1, 4), (1, 2, 3)$ and $(2, 2, 2)$ => Together contributing $10 $distinct triplets

Sum 7: $(1, 1, 5), (1, 2, 4), (1, 3, 3) $and $(2, 2, 3)$ => Together contributing $15$ distinct triplets.

Sum 8: $(1, 1, 6), (1, 2, 5), (1, 3, 4), (2, 3, 3)$ and$ (2, 4, 2) $==> Together contributing $21$ distinct triplets.

Sum 9: $(1, 2, 6), (1, 3, 5), (1, 4, 4), (2, 3, 4), (2, 5, 2)$ and $(3, 3, 3)$ ==> Together contributing $25$ distinct triplets.

Sum 10: $(1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4)$ and $(3, 3, 4) $==> Together contributing $27$ distinct triplets.

Therefore number of favorable cases $= 1+ 3 + 6 + 10 + 15 + 21 + 25 + 27 = 108.$

Therefore, probability $= \dfrac{108}{216} =\dfrac{1}{2}$
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

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

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 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.