3 Marks Question

Question 13 Marks

From the employees of a company, 5 persons are selected to represents them in the managing committee of the company particulars of five persons are as follows:

S.No.	Name	Sex	Age in years
1	Harish	M	30
2	Rohan	M	33
3	Sheetal	F	46
4	Alice	F	28
5	Salim	M	41

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

Answer

Here total number of persons = 5
One spokesperson is selected out of 5 persons in = 5 ways.
Let A be the event that the selected person is male
and B be the event that the selected person is over 35 years.
There are 3 male and one person can be selected in 3 ways.
$\therefore$ $P(A)={\tfrac35}$
There are 2 persons who are over 35 years.
So, out of themone person can be selected in 2 ways.
$\therefore$ $P(B)={\frac25}$
Since there is 1 male who isover 35 years.
so, $P(A\cap B)={\frac15}$
$\therefore$ P (that the spokesperson is either male or over 35 years)
= $P(A\cup B)$
$= P(A)+P(B)-P(A\cap B)$
$= \frac{3}{5} + \frac{2}{5} - \frac{1}{5} = \frac{{3 + 2 - 1}}{5}$
$= \frac{4}{5}$

View full question & answer→

Question 23 Marks

Out of $100$ students, two sections of $40$ and $60$ are formed. If you and your friend are among the $100$ students, what is the probability that,

you both enter the same section?
you both enter the different sections?

Answer

Given, total number of students $= 100.$
Out of these students two students can be chosen in ${{}^{100}C_2}$ ways.

If both of us belong to the same section
i.e., either both of us belong to the section of $40$ students or to the section of $60$ students.
$\therefore$ Number of total favourable outcomes in which both of us $($i.e., $2$ students$)$
belong to either section of $40$ students or section of $60$ students $= {{}^{40}C_2} + \style{}{{}^{60}C_2}$.
$\therefore$ Probability $($both of us belong to same section$)
= {\frac{{}^{40}C_2+{}^{60}C_2}{{}^{100}C_2}=\frac{{\frac{40!}{2!\;.(40-2)!}}+{\frac{60!}{2!\;.\;(60-2)!}}}{\frac{100!}{2!\;.(100-2)!}}}$
${=\frac{40\times39+60\times59}2\times\frac2{100\times99}}$${=\frac{4\times39+6\times59}{10\times99}=\frac{17}{33}}$
$P ($both students are in different sections$)$
$= 1 - \frac{{17}}{{33}} = \frac{{16}}{{33}}$

View full question & answer→

Question 33 Marks

The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e. from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Answer

There are total 10 digits from 0 to 9.
Since the digits cannot be repeated.
So the first place may be filled in 10 ways, second place in 9 ways, third place in 8 ways and fourth place in 7 ways.
$\therefore$ Number of possible outcomes $ = 10 \times 9 \times 8 \times 7 = 5040$
The lock of suitcase can be opened in 1 way only
$\therefore$ Number of favourable cases = 1
Thus required probability $ = \frac{1}{{5040}}.$

View full question & answer→

Question 43 Marks

A fair coin is tossed four times, and a person win Re $1$ for each head, and lose Rs. $1.50$ for each tail that turns up. Form the sample space, calculate how many different amounts of money he can have after four tosses and the probability of having each of these amounts.

Answer

Here a coin is tossed four times. So number of elements in the sample space (S) will be $2^4 = 16. n(S) = 16.$
The sample space,
$S = {HHHH, HHHT, HHTH, HTHH, HTTH, HTHT, HHTT, HTTT, THHH, THHT, THTH, TTHH, TTTH, TTHT, THTT, TTTT}$
Amounts:
(i) When $4$ heads turns up $=\operatorname{Rs}(1+1+1+1)=$ Rs. 4. i.e.,Person wins Rs. $4$
(ii) When $3$ heads and $1$ tail turns $u p=\operatorname{Rs}(1+1+1-1.50=$ Rs. 1.50. i.e.,Person wins Rs. $1.50$
(iii) When $2$ heads and $2$ tails turns up $=\operatorname{Rs}(1+1-1.50-1.50)=-$ Rs. 1. i.e., Person losesRs. $1$
(iv) When $1$ head and $3$ tails turns up $=\operatorname{Rs}(1-1.50-1.50-1.50)=-R s 3.50$. i.e., Person losesRs. $3.50$
(v) When4 tails turns up $=\operatorname{Rs}(-1.50-1.50-1.50-1.50)=-$ Rs 6 . i.e., Person loses Rs. $6$
Let the events for which the person wins Rs $4$ , wins Rs $1.50 $, loses Re $1$ , loses Rs $3.50$ and loses Rs $6$
Let the events for which the person wins Rs $4$, wins Rs $1.50$, loses Re$1$, loses Rs $3.50$ and loses Rs $6$
be denoted by $E_1, E_2, E_3, E_4$ and $E_5.$
i.e., $E_1 = {HHHH}, E_2 = {HHHT, HHTH, HTHH, THHH} E_3={HHTT,HTHT,HTTH,THTH,THHT,TTHH}$
$E_4 = {HTTT, TTTH, THTT, TTHT}, E_5 = {TTTT}$
Here, $n(E_1) = 1, n(E_2) = 4, n(E_3) = 6, n(E_4) = 4$ and $n(E_5) = 1.$
Hence,$\style{font-size:26px}{P(E_1)=\frac{n({E_1)}}{n(S)}=\frac1{16}}$,$\style{font-size:26px}{P(E_2)=\frac{n({E_2)}}{n(S)}=\frac4{16}=\frac14}$
$\style{font-size:26px}{P(E_3)=\frac{n({E_3)}}{n(S)}=\frac6{16}=\frac38}$
$\style{font-size:26px}{P(E_4)=\frac{n({E_4)}}{n(S)}=\frac4{16}=\frac14}$
and $\style{font-size:26px}{P(E_5)=\frac{n({E_5)}}{n(S)}=\frac1{16}=\frac1{16}}$

View full question & answer→

Question 53 Marks

A letter is chosen at random from the word ASSASSINATION find the probability that letter is a consonant.

Answer

In the word ASSASSINATION there are 6 vowels and 7 consonants.
Total 13 letters are there. So, total number of outcomes =13
Out of 7 consonants, 1 consonant can be selected in 7 ways.
So, total number of favourable outcomes = 7
$\therefore$ P( 1 consonant selected) $=\frac 7{13}$

View full question & answer→

Question 63 Marks

A letter is chosen at random from the word ASSASSINATION find the probability that letter is a vowel.

Answer

There are 13 letters in the word ASSASSINATION of which 6 are vowels and 7 are consonants.
In all there are 13 letters. So, total possible outcomes = 13
Out of 6 vowels, 1 vowel can be selected in 6 ways.
So, number of favourable outcomes = 6
$\therefore$ P( 1 vowel selected) $=\frac 6{13}$

View full question & answer→

Question 73 Marks

If A, B, C are three events associated with a random experiment, prove that
$P(A\cup B\cup C) = P(A) + P(B) +P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C)$

Answer

Consider E = $B \cup C$.
So P ($A \cup B \cup C$) = P ($A ∪ E$)
= P(A) + P(E) - $P(A\cap E)$ ..................(1)
Now P(E) = $P(B\cup C)$ .......................(2)
Also $(A\cap E) = A\cap (B\cup C) = (A\cap B)\cup (A\cap C)$ [using distribution property of intersection of sets over the union].
Writing them as probability,
$P(A\cap E) = P(A\cap B) (A\cup C) - P[(A\cap B)\cap (A\cap C)]$
$P(A\cap B) + P(A\cap C) - P[A\cap B\cap C]$..........................(3)
using (2) and (3) in (1), we get
$P[A\cup B\cup C] = P(A) + P(B) + P(C) - P(B\cap C)- P(A\cap B) - P(A\cap C) + P(A\cap B\cap C)$
Hence Proved.

View full question & answer→

Question 83 Marks

Find the probability that when a hand of $7$ cards is drawn from a well shuffled deck of $52$ cards, it contains

all Kings
$3$ Kings
at least $3$ Kings

Answer

Total number of possible hands$=$Total number of outcomes $=\ ^{52}C_7$

Number of hands with $4$ Kings $=\ ^4C_4\; \times \;^{48}C_3 ($as there are $4$ kings and other $3$ cards must be chosen from the rest $48$ cards$)$
Hence $P ($a hand will have $4$ Kings$) = \frac {^4C_4\; \times \;^{48}C_3}{^{52}C_7} = \frac 1{7735}$
Number of hands with $3$ Kings and $4$ non$-$King cards $=\ ^4C_3\; \times \;^{48}C_4$
Therefore $P (3$ Kings$) = \frac {^4C_3\; \times \;^{48}C_4}{^{52}C_7} = \frac 9{1547} ($as $3$ cards out of $4$ kings and other $4$ cards must be chosen from the rest $48$ cards$)$
$P($at least $3$ King$) = P(3$ Kings or $4$ Kings$)$
$=P(3$ Kings$) + P(4$ Kings$)$
$= \frac 9{1547}+ \frac 1{7735} = \frac {46}{7735}$

View full question & answer→

Maths 3 Marks Question

Take a timed test