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 business maths probability - iii baye's theorem bayes theorem probability and probability distribution

A is known to tell the truth in $5$ cases out of $6$ and he states that a white ball was drawn from a bag containing $8$ black and $1$ white ball. The probability that the white ball was drawn, is

  1. $\displaystyle \frac { 7 }{ 13 } $
  2. $\displaystyle \frac { 5 }{ 13 } $
  3. $\displaystyle \frac { 9 }{ 13 } $

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

Let $W$ denote the event that $A$ draws a white ball and $T$ the event that $A$ speak truth.
In the usual notations, we are given that 
$\displaystyle P\left( W \right) =\frac { 1 }{ 9 } ,P\left( \frac { T }{ w }  \right) =\frac { 5 }{ 6 } $
so that $\displaystyle P\left( \overline { W }  \right) =1-\frac { 1 }{ 9 } =\frac { 8 }{ 9 } ,P\left( \frac { T }{ \overline { W }  }  \right) =1-\frac { 5 }{ 6 } =\frac { 1 }{ 6 } $.
Using Baye's theorem required probability is given by 
$\displaystyle P\left( \frac { W }{ T }  \right) =\frac { P\left( W\cap T \right)  }{ P\left( T \right)  } =\frac { P\left( W \right) P\left( \frac { T }{ w }  \right)  }{ P\left( W \right) P\left( \frac { T }{ w }  \right) +P\left( \overline { W }  \right) P\left( \frac { T }{ \overline { W }  }  \right)  } $
$\displaystyle =\frac { \dfrac { 1 }{ 9 } \times \dfrac { 5 }{ 6 }  }{ \dfrac { 1 }{ 9 } \times \dfrac { 5 }{ 6 } +\dfrac { 8 }{ 9 } \times \dfrac { 1 }{ 6 }  } =\frac { 5 }{ 13 } $

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

A box has four dice in it. Three of them are fair dice but the fourth one has the number five on all of its faces. A die is chosen at random from the box and is rolled three times and shows up the face five on all the three occasions. The chance that the die chosen was a rigged die, is

  1. $\displaystyle \frac {216}{217}$
  2. $\displaystyle \frac {215}{219}$
  3. $\displaystyle \frac {216}{219}$

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

Given, $A$ is the event which selects the rigged one and $B$ is the event which selects the fair one.
Let E is the event which shows 5 in all three times
Probability of event A for given event E, $P(A/E)=\dfrac{1}{4}(1)^3=\dfrac{1}{4}$ ($\dfrac{1}{4}$ in the equation is probability of selecting one dice among 4)
Probability of event B for given event E, $P(B/E)=\dfrac{3}{4}(\dfrac{1}{6})^3=\dfrac{1}{1152}$ (since probability of getting 5 in fair dice case=$\dfrac{1}{5}$)
By Baye's theorem,Probability of selecting the rigged case among both=$\dfrac{P(A/E)}{P(A/E)+P(B/E)}=\dfrac{(\dfrac{1}{4})}{(\dfrac{1}{4})+(\dfrac{1}{1152})}=\dfrac{216}{219}$

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

Suppose that of all used cars of a particular year 30% have bad brakes. You are considering buying a used car of that year. You take the car to a mechanic to have the brakes checked. The chance that the mechanic will give you the wrong report is 20%. Assuming that the car you take to the mechanic is selected at random from the population of cars of that year. The chance that the car's brakes are good, given that the mechanic says its brakes are good, is

  1. $\displaystyle \frac{28}{130}$
  2. $\displaystyle \frac{29}{31}$
  3. $\displaystyle \frac{37}{62}$
  4. $\displaystyle \frac{29}{62}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $30$% of the cars of bad brakes
$P(E _1)=70$%$=\dfrac7{10}$        $P(E _2)=30$%$=\dfrac3{10}$
$\Rightarrow P\left( \dfrac { { E } }{ { E } _{ 1 } }  \right) =0.2\times0.2=0.04$
$\Rightarrow P\left( \dfrac { { E } }{ { E } _{ 2 } }  \right) =0.8\times0.8=0.64$
$\therefore P\left( \dfrac { { E } }{ { E } _{ 1 } }  \right) =\dfrac { \dfrac { 7 }{ 10 } \times \dfrac { 2 }{ 10 } \times \dfrac { 2 }{ 10 }  }{ \dfrac { 7\times 4 }{ 1000 } +\dfrac { 3 }{ 10 } +\dfrac { 8 }{ 10 } +\dfrac { 8 }{ 10 }  } =\dfrac { 28 }{ 102+28 } =\dfrac { 28 }{ 130 } $
Hence, the answer is $\dfrac { 28 }{ 130}.$
Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

Box $I$ contains $5$ red and $4$ blue balls, while box $II$ contains $4$ red and $2$ blue balls. A fair die is thrown. If it turns up a multiple of $3$, a ball is drawn from the box $I$ else a ball is drawn from box $II$. Find the probability of the event ball drawn is from the box $I$ if it is blue.

  1. $\displaystyle \frac{1}{6}$
  2. $\displaystyle \frac{15}{19}$
  3. $\displaystyle \frac{4}{19}$
  4. $\displaystyle \frac{10}{27}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Box $I$ contains $:5$ red $+{4}$ blue
Box $II$ contains $:4$ red $+{2}$ blue
A ball is taken from Box $I$ if a multiple of $3$ comes up i.e, $3$ and $6.$

Ball is taken from Box $II$ when $1,2,4$ and $5$ turns up.

$\Rightarrow$ Event of picking up from Box $I=P(A _1)=\dfrac{2}{6}=\dfrac{1}{3}.$

$\Rightarrow$ Event of picking up from Box $II=P(A _2)=\dfrac{4}{6}=\dfrac{2}{3}.$

$\Rightarrow R=$ event of drawing a blue ball

$=P(A _1)P(\dfrac R{A _1})+P(A _2)P(\dfrac{R}A _2)$

$=\dfrac{1}{3}\times\dfrac{4}{9}+\dfrac{2}{3}\times\dfrac{2}{6}$

$=\dfrac{4}{27}+\dfrac{4}{18}$

$=\dfrac{10}{27}.$
Hence, the answer is $\dfrac{10}{27}.$

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

There are three different Urns, Urn-I, Urn-II and Urn-III containing 1 Blue, 2 Green, 2 Blue, 1 Green, 3 Blue, 3 Green balls respectively. If two Urns are randomly selected and a ball is drawn from each Urn and if the drawn balls are of different colours then the probability that chosen Urn was Urn-I and Urn-II is

  1. $\dfrac {1}{7}$
  2. $\dfrac {5}{13}$
  3. $\dfrac {5}{14}$
  4. $\dfrac {5}{7}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Required probability$\displaystyle =\dfrac {\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {1}{3}+\dfrac {2}{3}.\dfrac {2}{3}\right )}{\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {2}{3}.\dfrac {2}{3}\right )+\dfrac {1}{3}\left (\dfrac {2}{3}.\dfrac {3}{6}+\dfrac {1}{3}.\dfrac {3}{6}\right )+\dfrac {1}{3}\left (\dfrac {3}{6}.\dfrac {2}{3}+\dfrac {3}{6}.\dfrac {1}{3}\right )}$

$\displaystyle =\dfrac {\dfrac {5}{9}}{\dfrac {5}{9}+\dfrac {9}{18}+\dfrac {9}{18}}\\ =\dfrac {5}{14}$

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

There are two groups of subjects one of which consists of 5 science subjects and 3 engineering subjects and the other consists of 3 science and 5 engineering subjects. An unbaised die is cast. If number 3 or number 5 turns up, a subject is selected at random from the first group, other wise the subject is selected at random from the second group. Find the probability that an engineering subject is selected ultimately.

  1. $\displaystyle \frac{13}{24}$
  2. $\displaystyle \frac{1}{3}$
  3. $\displaystyle \frac{2}{3}$
  4. $\displaystyle \frac{11}{24}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let  $\displaystyle E _{1}$ be the event that a subject is selected from first group.
$\displaystyle E _{2}$ the event that a subject is selected from the second group.
$E$ be the event that an engineering subject is selected.
Now the probability that die shows $3$ or $5$  is
$\displaystyle P(E _1)=\frac{2}{6}=\frac{1}{3}$

$\displaystyle P\left ( E _{2} \right )=\frac{1}{3}=\frac{2}{3}.$
Now probability of choosing an engineering subject from first group is 
$\displaystyle P\left ( E|E _{1} \right )=$  $\displaystyle \frac{^{3}C _{1}}{^{8}C _{1}}=\frac{3}{8}$
Similarly, $\displaystyle P\left( E|E _{2} \right )=\frac{^{5}C _{1}}{^{8}C _{1}}=\frac{5}{8}$ 
Hence $\displaystyle P\left ( E \right )=P\left ( E _{1} \right )P\left ( E|E _{1} \right )+P\left ( E _{2}\right )P\left ( E|E _{2} \right )$
$\displaystyle =\frac{1}{3}.\frac{3}{8}+\frac{2}{3}.\frac{5}{8}$

$=\dfrac{13}{24}$ 

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is $10$. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains $ 1$ white and $9$ black balls is

  1. $\dfrac {14}{55}$
  2. $\dfrac {12}{55}$
  3. $\dfrac {2}{11}$
  4. $\dfrac {8}{55}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $E _i$ denote the event that the bag contains $i$ black and ($10-i$) white balls $(i=0, 1, 2, ...., 10)$. Let $A$ denote the event that the three balls drawn at random from the bag are black. We have
$P(E _i)=\dfrac {1}{11} (i=0, 1, 2, ...., 10)$
$P(A|E _i)=0$ for $i=0, 1, 2$
and $P(A|E _i)=\dfrac {^iC _3}{^{10}C _3}$ for $i\geq 3$
Now, by the total probability rule
$\displaystyle P(A)=\sum _{i=0}^{10}P(E _i)P(A|E _i)$
$=\frac {1}{11}\times \frac {1}{^{10}C _3}[^3C _3+^4C _4+....+^{10}C _3]$
But $^3C _3+^4C _3+^5C _3+....+^{10}C _3$
$=^4C _4+^4C _3+^5C _3+...+^{10}C _3$
$=^5C _4+^5C _3+^6C _3+....+^{10}C _3$
$=^6C _4+^6C _3+....+^{10}C _3=....=^{11}C _4$
Thus, $P(A)=\dfrac {^{11}C _4}{11\times ^{10}C _3}=\dfrac {1}{4}$
By the Bayes' rule
$P(E _9|A)=\dfrac {P(E _9)P(A|E _9)}{P(A)}=\dfrac {\dfrac {1}{11}\dfrac {(^9C _3)}{^{10}C _3}}{\dfrac {1}{4}}=\dfrac {14}{55}$.

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

A box contain $N$ coins, $m$ of which are fair are rest and biased. The probability of getting a head when a fair coin is tossed is $1/2$, while it is $2/3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is

  1. $\dfrac {9m}{8N+m}$
  2. $\dfrac {m}{8N+m}$
  3. $\dfrac {N}{8n+m}$
  4. $\dfrac {1}{m}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $E _1, E _2$ and $A$ denote the following events:
$E _1$: coin selected is fair
$E _2$: coin selected is biased
$A: $ the first toss results in a head and the second toss results in a tail.
$\displaystyle P(E _1)=\frac {m}{N}, P(E _2)=\frac {N-m}{N}$,
$\displaystyle P(A|E _1)=\frac {1}{2}\times \frac {1}{2}\times \frac {1}{4}, P(A|E _2)=\frac {2}{3}\times \frac {1}{3}=\frac {2}{9}$.
By Bayes' rule
$\displaystyle P(E _1|A)=\frac {P(E _1)P(A|E _1)}{P(E _1)P(A|E _1)+P(E _2)P(A|E _2)}=\frac {9m}{8N+m}$.

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

A bag contains $(2n+1)$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\displaystyle \frac{31}{42}$, then $n$ is equal to 

  1. $10$
  2. $11$
  3. $12$
  4. $13$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let ${ A } _{ 1 }$ denote the event that a coin having heads on both sides is chosen, and ${ A } _{ 2 }$ denote the vent that a fiar coin is chosen.
Let $E$ denote the vent that head occurs, Then
$\displaystyle P\left( { A } _{ 1 } \right) =\frac { n }{ 2n+1 } \Rightarrow P\left( { A } _{ 2 } \right) =\frac { n+1 }{ 2n+1 } $
Probability of occurrence of event $E$, if unfair coin was selected is $\displaystyle P\left( \frac { E }{ { A } _{ 1 } }  \right) =1$
Probability of occurrence of event $E$, if fair coin was selected is $\displaystyle P\left( \frac { E }{ { A } _{ 2 } }  \right) =\frac { 1 }{ 2 } $
$\because P\left( E \right) =P\left( { A } _{ 1 }\cap E \right) +P\left( { A } _{ 2 }\cap E \right) $
$\displaystyle \therefore P\left( E \right) =P\left( { A } _{ 1 } \right) P\left( \frac { E }{ { A } _{ 1 } }  \right) +P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \right) $
$\displaystyle \Rightarrow \frac { 31 }{ 42 } =\frac { n }{ 2n+1 } .1+\frac { n+1 }{ 2n+1 } .\frac { 1 }{ 2 } \Rightarrow \frac { 31 }{ 42 } =\frac { 3n+1 }{ 2\left( 2n+1 \right)  } \ \Rightarrow 124n+62=126n+42\Rightarrow 2n=20\Rightarrow n=10$

Multiple choice business maths probability - iii baye's theorem bayes theorem probability and probability distribution

The contents of urn I and II are as follows:
Urn I: 4 white and 5 black balls
Urn II: 3 white and 6 black balls
One urn is chosen at random and a ball is drawn and its colour is noted and replaced back to the urn. Again a ball is drawn from the same urn colour is noted and replaced. The process is repeated 4 times and as a result one ball of white colour and 3 of black colour are noted. Find the probability the chosen urn was I.

  1. $\displaystyle \frac{125}{287}$
  2. $\displaystyle \frac{64}{127}$
  3. $\displaystyle \frac{25}{287}$
  4. $\displaystyle \frac{79}{192}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Bayes' theorem, we compare the likelihood of the observed sequence of balls (1 white, 3 black) occurring in Urn I versus Urn II. The calculation involves binomial probabilities for each urn.

Multiple choice botany classical genetics incomplete dominance deviation from mendelism inheritance

The segregration of alleles is a random process and so there is a _______ chance of a gametes containing either allele.

  1. $25\%$
  2. $50\%$
  3. $75\%$
  4. $100\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Mendel"s law of segregation, alleles segregate from one another so that the gametes will have only one copy of the particular gene. In a diploid organism, there are two alleles for a particular gene, which segregate and only one of each allele can be present in a gamete, therefore, since there are two alleles for a gene, chances of a gamete containing either of the allele is 50%. 

So, the correct option  is '50%'.

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

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

  1. $0$
  2. $\displaystyle\frac { 1 }{ 3 } $
  3. $\displaystyle\frac { 1 }{ 12 } $
  4. $\displaystyle\frac { 1 }{ 36 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two dice are rolled, the number of outcomes is $36$.
The only even prime number is $2$.
Let $E$ be the event of getting an even prime number on each die.
$\therefore E=\left{ \left( 2,2 \right)  \right} $
$\Rightarrow P\left( E \right) =\displaystyle\frac { 1 }{ 36 } $
Therefore, the correct answer is (D).

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

Two cards are drawn from a single deck of $52$ cards one after the other. Find the probability of selecting a king from the first card and queen from the second card.

  1. $\dfrac{1}{26}$
  2. $\dfrac{4}{52}$
  3. $\dfrac{16}{663}$
  4. $\dfrac{4}{663}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Probability of selecting a king in the first card is $\dfrac{4}{52}$. (Since $4$ kings in $52$ cards).

Probability of selecting a queen from the second card after the first card is drawn out is $\dfrac{4}{51}$. (Since $4$ queens in left over $51$ cards.
Now probability of selecting a king from the first card and queen from the second card is $\dfrac{4}{52}\times \dfrac{4}{51}=\dfrac{4}{663}$.
Hence, option D is correct.