Tag: statistics and probability

Questions Related to statistics and probability

Find the probability of getting a total of $7$ or $11$ when a pair of dice is tossed.

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

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

  2. $\dfrac{7}{9}$

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

  4. $\dfrac{5}{9}$

Show answer
Correct Option: C
Explanation:
Total number of all possible outcomes, $n(S)=6\times 6=36$
Let A: be the event that the sum is $7$.
Hence the favourable outcomes are ${(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}$, i.e., $n(A)=6$
Hence the probability of getting a total of $7$. $P(E)=\dfrac {n(A)}{n(S)}=\dfrac {6}{36}$
Similarly, let B: be the event that the sum is $11$
Hence the favourable outcomes are {(5, 6), (6, 5)}, i.e., $n(E)=2$
Hence the probability of getting a total of 11 $P(B)=\dfrac {n(B)}{n(S)}=\dfrac {2}{36}$
Now, the probability of getting a total of $7$ or $11$ is
$P(A)+P(B)\\\implies =\dfrac 6{36}+\dfrac 2{36}=\dfrac 8{36}=\dfrac 29$

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 none of the posts 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. $\dfrac{1}{4}$

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

Show answer
Correct Option: C
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.
$\Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(A\cap C)-P(B\cap C)+P(A\cap B \cap C)$
$\Rightarrow P(A)=\dfrac{1}{3}$   $P(B)$  $P(C)=\dfrac{1}{2}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\left( \dfrac {1  }{  3} \times\dfrac{1}{4}\right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{2}  \right) -\left( \dfrac { 1 }{ 4 }\times\dfrac{1}{2}  \right)=\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{4} \dfrac{1}{2} \right)$
                                 $=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{24}$
                                 $=\dfrac{3}{4}$
$\Rightarrow P$ ( not getting selected for any of the post ) $=1-\dfrac{3}{4}$
                                                                                $=\dfrac{1}{4}$
Hence, the answer is $\dfrac{1}{4}.$

A number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either $3$ or $13$?

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

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

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

  3. $\dfrac{11}{27}$

  4. $\dfrac{9}{27}$

Show answer
Correct Option: A
Explanation:

Given: first $30$ natural numbers

To find: the probability of getting a multiple of either $3$ or $13$
According to the question, 
$n(S)=30$
Multiple of 3 in first 30 natural numbers are ${3, 6, 9, 12, 15, 18, 21, 24, 27, 30} = 10$
Multiple of $13$ in first $30$ natural numbers are ${13, 26}$
Hence the probability of getting a multiple of either $3$ or $13$ = probability of getting a multiple of 3 or probability of getting a multiple of 13
$\implies \dfrac {10}{30}+\dfrac 2{30}=\dfrac {12}{30}=\dfrac 25$
is the required probability.

If $P(A)=\dfrac {1}{8}$ and $P(B)=\dfrac {5}{8}$. Which of the following statement is/are not correct?

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

  1. $P(A\cup B)\leq \dfrac {3}{4}$

  2. $P(A\cap B)\leq \dfrac {1}{8}$

  3. $P(\overline A\cap B)\leq \dfrac {5}{8}$

  4. None of these

Show answer
Correct Option: A,B,C
Explanation:

Given that $P\left( A \right) =\cfrac { 1 }{ 8 } ,\quad P\left( B \right) =\cfrac { 5 }{ 8 } $

we know that $P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $
a. $P\left( A\cup B \right) =\cfrac { 3 }{ 4 } -P\left( A\cup B \right)$
 $ \ P\left( A\cup B \right) $ would be max when $P\left( A\cap B \right) =0$
so,  
$P\left( A\cup B \right) \quad \le \quad \cfrac { 3 }{ 4 } $
So option A is correct.

b. 
$P\left( A\cup B \right) \quad \le \quad min\left( P\left( A \right) ,P\left( B \right)  \right) \ $
In this case, $min\left( P\left( A \right) ,P\left( B \right)  \right) \quad =\quad \frac { 1 }{ 8 } $
So, $P\left( A\cup B \right) \quad \le \quad \cfrac { 1 }{ 8 } $
Option B is correct.

c
$P\left( \bar { A }  \right) =1-P\left( A \right) \ \qquad \quad \quad =1-\cfrac { 1 }{ 8 } \ \quad \quad \quad \quad \quad =\cfrac { 7 }{ 8 } \ P\left( A\cup B \right) \quad \le \quad min\left( P\left( A \right) ,P\left( B \right)  \right) $
So, 
$P\left( \bar { A } \cap B \right) \quad \le \quad \cfrac { 5 }{ 8 } $
Option C is correct.

Addition Theorem of Probability states that for any two events $A$ and $B$,

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

  1. $P(A$ $\cup$ $B) = P(A) + P(B)$

  2. $P(A$ $\cup$ $B) = P(A) + P(B) + P(A$ $\cap$ $B)$

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

  4. $P(A$ $\cup$ $B) = P(A) \times P(B)$

Show answer
Correct Option: C
Explanation:
For Addition theorem,
If $A$ and $B$ are any two events associated with a random experiment, then
$\Rightarrow P(A\cup B) = P(A) + P(B)- P(A\cap B)$
Hence, the answer is $P(A\cup B) = P(A) + P(B)- P(A\cap B).$

A, B and C in order toss a coin. the first one to throw a head wins. If A starts to toss, then 

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

  1. $P(A)=\dfrac{4}{7}$

  2. $P(B)=\dfrac{2}{7}$

  3. $P(C)=\dfrac{1}{7}$

  4. $P(C)=\dfrac{2}{7}$

Show answer
Correct Option: B
Explanation:
P(A)=H+TTTH+TTTTTTH+.....
        =$1/2+{1 (1/2 })^{ 4 }+{ (1/2 })^{ 7 }+....$
        $=\frac { 1/2 }{ 1-1/8 } =4/7$
P(B)=TH+TTTTH+TTTTTTTH+.....
        =${ (1/2 })^{ 2 }+{ (1/2 })^{ 5 }+....$
        $=\frac { 1/4 }{ 1-1/8 } =2/7$
P(C)=TTH+TTTTTH+TTTTTTTTH+.....
        =${ (1/2 })^{ 3 }+{ (1/2 })^{ 6 }+....$
        $=\frac { 1/8 }{ 1-1/8 } =1/7$

A bag contain $5$ balls of unknown colors. A ball is drawn at random from it and is found to be white. The probability that bag contains only white ball is

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

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

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

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

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

Show answer
Correct Option: D
Explanation:
Let $Q$ be the event that the drawn balls are white

$A$ be the event that the bag contains $1$ white ball
$B$ be the event that the bag contains $2$ white balls
$C$ be the event that the bag contains $3$ white balls
$D$ be the event that the bag contains $4$ white balls
$E$ be the event that the bag contains $5$ white balls

$P(A)=1/5,\ P(B)=1/5,\ P(C)=1/5,\ P(D)=1/5,\ P(E)=1/5$
$P(Q/A)=\dfrac {^1C _1}{^5C _1}=\dfrac {1}{5}$
$P(Q/B)=\dfrac {^2C _1}{^5C _1}=\dfrac {2}{5}$
$P(Q/C)=\dfrac {^3C _1}{^5C _1}=\dfrac {3}{5}$
$P(Q/D)=\dfrac {^4C _1}{^5C _1}=\dfrac {4}{5}$
$P(Q/E)=\dfrac {^5C _1}{^5C _1}=1$

using Baye's thorem,

$P(A/Q)=\dfrac {P(E)\ P(Q/E)}{P(A)\ P(Q/A)+P(B)\ P(Q/B)+P(C)\ P(Q/C)+P(D)\ P(Q/D)+P(E)\ P(Q/E)}$

$=\dfrac {1/5. 1}{1/5. 1/5+1/5.2/5+1/5.3/5 +1/5.4/5+1/5.1.}$

$=\dfrac {1/5}{1/5 (1/5)+2/5+3/5+4/5+1}$

$=\dfrac {1}{\dfrac {1+2+3+4+5}{5}}=\dfrac {5}{1+2+3+4+5}=\dfrac {5}{15}=\dfrac {1}{3}$

n men and n women are seated at round table in random order. The probability that they can be divided into n non-interrecting pairs so that each pair consists of a man and a women is

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

  1. 1/2n

  2. $2(2^n-1)/^{2n}C _n$

  3. $2n/^{2n}C _n$

  4. $1/(^nC _n)^2$

Show answer
Correct Option: B
Explanation:

The women can be seated in $^{2n}C _n$ ways. Let A denote the event that pairs occupy the seats (1, 2), (3, 4), ...., (2n-1, 2n)
and B denote the event that pairs occupy the seats (2, 3), (4, 5), (6, 7), ...., (2n-2, 2n-1), (2n-1).
The number of cases favourable to A (B) is $2^n$. (For each man in the pair there are two choices.)
However, there are just two cases common to A and B. One is the case $\left {(M, W), (M, W), ...(M, W)\right }$ of A(B) and $\left {(W, M), (W, M), ..., (W, M)\right }$ of B(A).
Therefore, $P(A\cup B)=P(A)+P(B)-P(A\cup B)$
$\displaystyle =\frac {2^n+2^n-2}{^{2n}C _n}=\frac {2^{n+1}-2}{^{2n}C _n}$.

A is interviewed for $3$ posts. There are $4$ candidates for post $1, 3$ for second post and $5$ 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. $\dfrac{3}{5}$

Show answer
Correct Option: D
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.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A)=\dfrac{1}{4}$    $P(B)=\dfrac{1}{3}$     $P(C)=\dfrac{1}{5}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{3} \right)-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{5} \right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{5} \right)+\left( \dfrac { 1 }{ 34}\times\dfrac{1}{3}\times\dfrac{1}{5}\right)$
                                 $=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{15}+\dfrac{1}{60}$

                                 $=\dfrac{15+20+12-5-3-4+1}{60}$
                                 $=\dfrac{3}{5}.$
Hence, the answer is $\dfrac{3}{5}.$

$A$ is interviewed for $3$ posts. There are $4$ candidates for post $1, 3$ for second post and $5$ for post No. three. the probability of $A$'s being selected for none of the posts is:

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

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

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

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

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

Show answer
Correct Option: B
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.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{3} \right)-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{5} \right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{5} \right)+\left( \dfrac { 1 }{ 34}\times\dfrac{1}{3}\times\dfrac{1}{5}\right)$
                                 $=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{15}+\dfrac{1}{60}$
                                 $=\dfrac{3}{5}.$
$\Rightarrow P$ ( not getting selected for none of the post ) $=1-\dfrac{3}{5}$
                                                                                   $=\dfrac{2}{5}$
Hence, the answer is $\dfrac{2}{5}.$