Tag: statistics and probability

Questions Related to 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 $\displaystyle P(A \cup B)$ is

  1. $\displaystyle \frac{2}{5}$

  2. $\displaystyle \frac{3}{5}$

  3. 0

  4. 1


Correct Option: D
Explanation:

Here, $\displaystyle A=\left{ 4,5,6\right} $
and  $\displaystyle B=\left{ 1,2,3,4,\right} $
We have $\displaystyle A\cup B=\left{1,2,3,4,5,6\right} =S$
Where S is the sample space of the experiment of throwing a die.
$\displaystyle P(S) =1$ since it is a sure event.Hence $\displaystyle P(A\cup B)=1$

Twenty bags or sugar, each marked 10 kg actually give the following data:

The weight of a bag (in kg) No. of bags
     $9.5-9.8$          1
     $9.8-9.9$          2
     $9.9-10.0$          5
     $10.0-10.1$         12

The lower limits of the classes are inclusive and the upper limits are exclusive.

What is the probability that the bag selected at random (without any reference) weighs 10 kg or more?

  1. $0.6$

  2. $0.8$

  3. $0.5$

  4. $0.3$


Correct Option: A
Explanation:

Total number of bags = $12 + 5 + 2 + 1 = 20$
Bags which weigh more than $10 kg = 12$
Thus Probability (bag weighs more than 10 kg) = $\dfrac{12}{20} = 0.6$

In an experiment, there are exactly three elementary events. The probability of two of them are $\displaystyle\frac{2}{7}$ and $\displaystyle\frac{1}{7}$. What is the probability of third event?

  1. $\displaystyle\frac{4}{7}$

  2. $\displaystyle\frac{3}{7}$

  3. $\displaystyle\frac{2}{7}$

  4. $\displaystyle\frac{1}{7}$


Correct Option: A
Explanation:

We know that the sum of all the elementary events = 1
$\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7} + P _3 = 1$ 

$\Rightarrow \quad P _3 = 1 - \left(\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7}\right) = \displaystyle\frac{4}{7}$ 

If $n(A)=6,n(B)=8$ and $n(A\cup B)=12$, then $n(A\cap B)=$

  1. $6$

  2. $2$

  3. $8$

  4. $12$


Correct Option: B
Explanation:
We know that $n(A)+n(B)=n(A\cup B)+n(A\cap B)$
$n(A)=6,$ $n(B)=8$ and $n(A\cup B)=12$
$n(A)+n(B)-n(A\cup B)=n(A\cap B)$
$n(A\cap B)=6+8-12=2$

Let A be a set of $4$ elements. From the set of all functions from A to A, the probability that it is an into function is?

  1. $\dfrac{3}{32}$

  2. $0$

  3. $\dfrac{29}{32}$

  4. $1$


Correct Option: A
Explanation:
Let set A be of n element and from A to A then no of onto element is n! and total function are $n^{2}$

$P(E)=\dfrac{n!}{n}=\dfrac{4!}{4\times 4\times 4\times 4}$

$=\dfrac{4\times 3\times 2}{4\times 4\times 4\times }=\dfrac{3}{32}$

From a well shuffled standard pack of $52$ playing cards, one card is drawn. What is the probability that it is either a King of hearts or a Queen of diamonds.

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

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

  3. $\dfrac {1}{8}$

  4. $\dfrac {1}{26}$


Correct Option: D
Explanation:

No. of Kings of heart $=$ 1, No. of Queen of diamonds $=$1
Probability of king of hearts or queen of diamonds $ = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$

If a coin is tossed twice, then the events 'occurrence of one head',  'occurrence of $2$ heads' and 'occurrence of no head' are -

  1. Independent

  2. Equally likely

  3. Not equally likely

  4. Both (A) and (B)


Correct Option: C
Explanation:
Possible outcomes when a coin is tossed $=$ H or T
Possible outcomes when a coin is tossed twice$=$ HH,TT,TH,HT
$=4$ outcomes
P(occurrence of $1H$) $=$ TH,HT
$=2/4=1/2$
P(occurrence of $2H$) $=$ HH
$=1/4$
P(occurrence of no H) $=$ TT
$=1/4$
Therefore, the probability of occurrence of no head and occurrence of no head and occurrence of two head is same but probability of occurrence of one Head is $1/2$. Thus the events are:-
Not equally likely.

If events A and B are independent and P(A) $=$ 0.15, P(A $\cup $ B) $=$ 0.45, Then P(B) $=$ ..............

  1. $\displaystyle \frac{6}{13}$

  2. $\displaystyle \frac{6}{17}$

  3. $0.315$

  4. $0.352$


Correct Option: B,D
Explanation:
Events A and B are independent then $P\left( A\bigcap { B }  \right) =0$
$P\left( A\bigcup { B }  \right) =P(A)+P(B)-P(A).P(B)$
Given: $P(A)=0.15$
$P\left( A\bigcup { B }  \right) =0.45$
$\therefore P(B)=P\left( A\bigcup { B }  \right) -P(A)+P\left( A \right) P(B)$
$P(B)=0.45-0.15+1.15P(B)$
$P(B)=0.3+0.15P(B)$
$P(B)(1-0.15)=0.3$
$P(B)(0.85)=0.3$
$P(B)=\cfrac { 0.3 }{ 0.85 } =0.352=\cfrac { 6 }{ 17 } $

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


Correct Option: A
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

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. $1$


Correct Option: D
Explanation:
$A \equiv \left\{4, 5, 6\right\}$
$B \equiv \left\{1, 2, 3, 4\right\}$
$\therefore \ A\cup B\equiv \left\{1, 2, 3, 4, 5, 6\right\}$
$\therefore \ P(A\cup B)=\dfrac {6}{6}=1$