Tag: statistics and probability

Questions Related to statistics and probability

An institute organised a fete and ${1}/{5}$ of the girls and ${1}/{8}$ of the boys participated in the same. What fraction of the total number of students took part in the fete?

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

  1. ${2}/{13}$

  2. ${13}/{40}$

  3. Data inadequate

  4. None of these

Show answer
Correct Option: C
Explanation:
Ans is "Data inadequate".

Suppose ther are 18 students (10 girls, 8 boys).

1/5 of 10=2 (girls).

1/8 of 8=1 (boy).

Total participation=2+1=3.

Total students=10+8=18.

Answer = 3/18 = 1/6.

Answer cannot be determined until we have the "Ratio of girls to boys".

For each different number of boys and girls, we get different answers. So we cannot obtain the required answers without knowing the ratio of boys and girls.

A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband`s selection is $\dfrac {1}{7}$ and that of wife's selection is $\dfrac {1}{5}$. What is the probability that only one of them will be selected?

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

  1. $\dfrac {1}{7}$

  2. $\dfrac {2}{7}$

  3. $\dfrac {3}{7}$

  4. $\dfrac {4}{7}$

Show answer
Correct Option: B
Explanation:

There would be three cases.

1. Husband is selected while wife doesn't get selected.
2. Wife is selected while husband doesn't get selected.
3. Both of them get selected.

Probability of husband not to get selected $=\dfrac{6}{7}$
Probability of wife not to get selected $=\dfrac{4}{5}$
Probability only one to get selected $=\dfrac{6}{7} \times \dfrac{1}{5} +\dfrac{1}{7} \times \dfrac{4}{5} $
$=\dfrac{10}{35} =\dfrac{2}{7}$

A number $x$ is selected from first $100$ natural numbers. Find the probability that $x$ satisfies the condition $x+ \dfrac{100}{x} >50$

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

  1. $\dfrac{55}{100}$

  2. $\dfrac{45}{100}$

  3. $1$

  4. $\dfrac{50}{100}$

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Correct Option: A
Explanation:

$x+\dfrac {100}x>50\\implies x^2 +100>50x\\implies x^2-50x+100+(25)^2-(25)^2>0\\implies x^2+(25)^2-50x-525>0\\implies (x-25)^2>525\\implies x-25>\pm\sqrt{525}\\implies x>25+22.91, x<25-22.91\\implies x>47.91, x<2.09\\implies x\ge 48, x\le 2$

As we have to select from 1 to 100 and x is greater than or equal to 48. Hence we have to select from 49 to 100 which is 53 numbers and favorable cases are 1, 2.
Total favorable case = 53+2=55.
Total number of cases = 100
Hence the required probability, $=\dfrac {55}{100}$

$A$ speaks the truth in $60\%$ cases and $B$ in $70\%$ cases. The probability that they will say the same thing while describing a single event is:

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

  1. $0.56$

  2. $0.54$

  3. $0.38$

  4. $0.94$

Show answer
Correct Option: B
Explanation:

Both will speak the same thing means both can say truth or both can say false

The probability of truth for $A$ is $0.6$ and for $B$ is $0.7$
The probability of false for $A$ is $0.4$ and for $B$ is $0.3$
The probability that both say same thing is $0.6 \times 0.7 + 0.4 \times 0.3 = 0.42+0.12=0.54$
Therefore option $B$ is correct

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

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

  1. $\dfrac{4}{9}$

  2. $\dfrac{1}{3}$

  3. $\dfrac{7}{18}$

  4. None of these

Show answer
Correct Option: D
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.

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

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

  1. $\dfrac{5}{36}$

  2. $\dfrac{1}{9}$

  3. $\dfrac{2}{9}$

  4. $\dfrac{19}{36}$

Show answer
Correct Option: C
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.

If A and B are mutually exclusive events such that $P(A)=\frac{3}{5}$ and $ P(B)=\frac{1}{5}$, then find $P(A \cup B)$. 

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

  1. $\dfrac{2}{5}$

  2. $\dfrac{3}{5}$

  3. $\dfrac{4}{5}$

  4. None of these

Show answer
Correct Option: C
Explanation:

$A\quad and\quad B\quad are\quad mutually\quad exclusive\quad events\quad than\quad P(A\cap B)=0\ P(A)=\frac { 3 }{ 5 } \quad and\quad P(B)=\frac { 1 }{ 5 } \ P(A\cup B)=P(A)+P(B)-P(A\cap B)\ \qquad \qquad =\frac { 3 }{ 5 } +\frac { 1 }{ 5 } -0=\frac { 4 }{ 5 } $

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?

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

  1. $1/2$

  2. $1/4$

  3. $2/3$

  4. $1/6$

Show answer
Correct Option: C
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, $P(\cup B)=P(A)+P(B)-P(A)(P(B)=\dfrac 12+\dfrac 13-\dfrac 12\times \dfrac 13=\dfrac {3+2-1}6=\dfrac 46=\dfrac 23$

If $A$ and $B$ are two events such that $P(A\cup B)=\cfrac { 3 }{ 4 } ,P(A\cap B)=\cfrac { 1 }{ 4 } ,P(\bar { A } )=\cfrac { 2 }{ 3 } $ where $\bar { A } $ is the complement of $A$, then what is $P(B)$ equal to?

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

  1. $\dfrac{1}{3}$

  2. $\dfrac{2}{3}$

  3. $\dfrac{1}{9}$

  4. $\dfrac{2}{9}$

Show answer
Correct Option: B
Explanation:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\ \Rightarrow P(A\cup B)=1-P(\overset { \ _  }{ A } )+P(B)-P(A\cap B)\ \Rightarrow \dfrac { 3 }{ 4 } =1-\dfrac { 2 }{ 3 } +P(B)-\dfrac { 1 }{ 4 } \ \Rightarrow P(B)=\dfrac { 2 }{ 3 } $
Hence, option B is correct.

A is interviewed for $3$ posts. There are $3$ candidates for post $1,4$ for second post and $2$ for post No. three. The probability of A's being selected for at least one post is:

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

  1. $\dfrac{3}{4}$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{3}$

  4. $\dfrac{1}{5}$

Show answer
Correct Option: A
Explanation:
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A)=\dfrac{1}{3}$    $P(B)=\dfrac{1}{4}$     $P(C)=\dfrac{1}{2}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\left( \dfrac { 1 }{ 3 }\times \dfrac{1}{4} \right)-\left( \dfrac { 1 }{ 3 }\times \dfrac{1}{2} \right)-\left( \dfrac { 1 }{ 4 }\times\dfrac{1}{2} \right)+\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{4}\times\dfrac{1}{2}\right)$
                                 $=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{24}$
                                 $=\dfrac{8+6+12-2-4-3+1}{24}$
                                 $=\dfrac{3}{4}.$
Hence, the answer is $\dfrac{3}{4}.$