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Questions Related to business maths

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

A card is randomly drawn from a well shuffled pack of $52$ playing cards. The probability that it is a club or numbered $5$ is 

  1. Not $(0.4 + 0.3)$
  2. $=$ $0.4 + 0.3$
  3. $=$$ 0.4 - 0.3$

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

One card is both a 5 and a Club.
$\therefore$ Club or numbered 5 are not mutually exclusive.
P(club or numbered 5) $\neq$ 0.4 + 0.3

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

Assume that the birth of a boy or girl to a couple to be equally likely,mutually exclusive exhaustive and independent of the other children in the family for a couple having $6$ children the probability that their 'three oldest are boy'is

  1. $\dfrac{{20}}{{64}}$
  2. $\dfrac{{1}}{{64}}$
  3. $\dfrac{2}{{64}}$
  4. $\dfrac{8}{{64}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that the birth of a boy or girl to a couple to be equally likely, 
$(1)$ mutually exclusive
$(2)$ exhaustive
$(3)$ independent
$\Rightarrow$ One event does not affect the other 
$P(E)=P(B).P(B).P(B), P(B\ or\ G). P(B\ or\ G)$
$P(B)=$ probability of boy is $\dfrac{1}{2}$
$P(G)=$ probability of girl is $\dfrac{1}{2}$
$P(E)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times 2$ either boy or girls
$=\dfrac{2}{64}$
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

If  $A$ and $B$ are two independent events such that $P\left({A}^{\prime}\right)=0.7,  P\left({B}^{\prime}\right)=p$ and $P\left( A\cup B \right) =0.8,$ then the value of $p$ is 

  1. $1$
  2. $0.2$
  3. $0.7$
  4. $0.3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$P(A')=0.7\Rightarrow P(A)=1-0.7=0.3$

$P(B)=1-p$ 

$P(A\cup B)=0.8$

$P(A\cap B)=P(A)\times P(B)=0.3(1-p)$ as they are independent events

$\therefore P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$0.8=0.3+1-p-0.3(1-p)$

$0.8=0.3+1-p-0.3+0.3p$

$0.8=1-0.7p\Rightarrow 0.7p=0.2$

$p=\dfrac 27\approx0.3$

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

One card is drawn from a pack of $52$ cards. The probability that the card picked is either a spade or a king.

  1. $ \displaystyle \frac{1}{26} $
  2. $ \displaystyle \frac{3}{26} $
  3. $ \displaystyle \frac{4}{13} $
  4. $ \displaystyle \frac{1}{9} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $A:$ event of drawing a king
$\displaystyle \Rightarrow P(A)=\dfrac{4}{52}=\dfrac{1}{13}$
$B:$ event of drawing a spade $\displaystyle =\dfrac{13}{52}=\dfrac{1}{4}$
$\Rightarrow \displaystyle P\left( A\cap B \right) =\dfrac { 1 }{ 52 } $
$\displaystyle \therefore P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) =\dfrac { 1 }{ 13 } +\dfrac { 1 }{ 4 } -\dfrac { 1 }{ 52 } =\dfrac { 4 }{ 13 } $
Multiple choice business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

In a survey conducted among 400 students of X standard in Pune district, 187 offered to join Science faculty after X std. and 125 students offered to join Commerce faculty after X, If a student is selected at random from this group. Find the probability that student prefers Science or Commerce faculty.

  1. $\displaystyle \frac{39}{50}$
  2. $\displaystyle \frac{4}{5}$
  3. $\displaystyle \frac{41}{50}$
  4. $\displaystyle \frac{43}{50}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total number of students $= 400$
No. of science students $= 187$
No. of commerce students $ = 125$
Thus, Probabilty of the student being either science or commerce = $\dfrac{187}{400} + \dfrac{125}{400}$
= $\dfrac{312}{400}$
= $\dfrac{39}{50}$

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

An integer is chosen at random from the first two hundred digits.What is the probability that the integer chosen is divisible by 6 or 8 ?

  1. $\displaystyle\frac{29}{100}.$
  2. $\displaystyle\frac{1}{4}.$
  3. $\displaystyle\frac{1}{8}.$
  4. $\displaystyle\frac{21}{100}.$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let A=the integer is divisible by 6 
$A={6,12,18,....198}$
$198=6+(n-1)6$
$\Rightarrow n =33$
So, $P(A)=\dfrac{33}{200}$

B=the integer is divisible by 8
$B={8,16,24,....200}$
$200=8+(n-1)8$
$\Rightarrow n=25$
So, $P(B)=\dfrac{25}{200}$

Both are divisible by 6 and 8 both 
$A\cap B=24,48,.....192$
$192=24+(n-1)24$
$\Rightarrow n=8$
So, $P(A\cap B)=\dfrac{8}{200}$

$\displaystyle \therefore P\left ( A\cup B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( AB \right )$
$\displaystyle=\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{1}{4}.$

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

A random variable $X$ has the probability distribution:

$x$ 1 2 3 4 5 6 7 8
$P(X=x)$ 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

For the events $E=\left {x|x \text {is prime}\right }$ and $F=\left {x|x < 4\right }$, the probability $P(E\cup F)$ is

  1. $0.35$
  2. $0.77$
  3. $0.87$
  4. $0.50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using given table, $P(E)=0.62, P(F)=0.5, P(E\cap F)=0.35$.
$\therefore P(E\cup F)=P(E)+P(F)-P(E\cap F)=0.62+0.5-0.35=.77$

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