Quantitative Aptitude

Probability

1,860 Questions
All Sample Space Setup 558 Permutation & Combination Counting 288 Multiplication Theorem (AND) 234 Addition Theorem (OR) 151 At Least One Probability 134 Complement Rule (P(Not A)) 130 Drawing Without Replacement 126 Conditional Probability 75 Bayes' Theorem 69 Drawing With Replacement 49 Success and Failure Rates 46

Probability Questions

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

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 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.

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

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

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

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.