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MATHS 2 Marks Questions

Probability · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 14 questions.

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Question 12 Marks
Three letters are written to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
Answer
Given that each envelope contains exactly one letter.
$\therefore$ Number of ways in which three letters can be inserted randomly into three envelopes = 3! = 6
Out of three letters insertion of one letter into proper envelope can be done in $\style{font-size:28px}{{}^3C_1=3}$ ways, and insertion of other two letters into wrong envelope can be done in one way.
$\therefore$ Number of ways in which one letter can be inserted into proper envelope and other two in wrong envelopes =$\style{font-size:28px}{{}^3C_1\times1=3}$
Also, number of ways in which two letters can be inserted into proper envelopes = 1
Number of ways in which atleast one letter is in proper envelope = 3 + 1 = 4
Thus the required probability that atleast one letter is in its proper envelope = $\style{font-size:24px}{\frac46=\frac23}$
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Question 22 Marks
4 cards are drawn from a well shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?
Answer
From a pack of 52 cards, 4 cards can be drawn in $^{52}{C_4}$ ways.
There are 13 cards of diamond and 18 cards of spades
Now 3 cards of diamond out of 18 cards of diamond can be drawn in $^{13}{C_3}$ ways and 1 card of spade out of 13 cards of spade can be drawn in $^{13}{C_1}$ ways.
Thus the probability of obtaining 3 diamond and 1 spade card
$= \frac{{^{13}{C_3}{ \times ^{13}}{C_1}}}{{^{52}{C_4}}}$
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Question 32 Marks
A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that at least one will be green?
Answer
Total marbles in box = 10 + 20 + 30 = 60
We have to draw 5 marbles out of 60 marbles and this can be done in $^{60}{C_5}$ ways.
There are 10 red marbles and 20 blue marbles.
We have to draw 5 marbles out of these 30 marbles ( for now green marbles are excluded), then none of the drawn marbles will be green and that can be done in $^{30}{C_5}$ ways.
Thus the probability that at least one drawn marbles will be green= 1- P(no green marble out of 5) $ = 1 - \frac{{^{30}{C_5}}}{{^{60}{C_5}}}$
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Question 42 Marks
A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that all will be blue?
Answer
Total marbles = 10 + 20 + 30 = 60
We have to draw 5 marbles out of 60 marbles and this can be done in $^{60}{C_5}$ ways.
There are 20 blue marbles.
We have to draw 5 marbles out of 20 blue marbles and that can be done in $^{20}{C_5}$ ways.
Thus the probability that all drawn marbles will be blue $= \frac{{^{20}{C_5}}}{{^{60}{C_5}}}$
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Question 52 Marks
There are four men and, six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Answer
Here total members in the council = 4 + 6 = 10
One member is selected out of 10 members
$\therefore \;n(S){ = ^{10}}{C_1} = 10$
Let A be the event that the member is a woman.
$n(A){ = ^6}{C_1} = 6$
Thus $P(A) = \frac{{n(A)}}{{n(S)}} = \frac{6}{{10}} = \frac{3}{5}$
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Question 62 Marks
A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is 12.
Answer
The coin with 1 marked on one face and 6 on the other face.
The coin and die are tossed together.
$\therefore$ S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
$\Rightarrow$ smpale space n(S) = 12
Let B be the event having a sum of the numbers is 12
therefore, B = {(6, 6)}
$\Rightarrow$ n (B) = 1
Thus, $P(B) = \frac{1}{{12}}$
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Question 72 Marks
A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is 3.
Answer
The coin with 1 marked on one face and 6 on the other face.
The coin and die are tossed together.
$\therefore$ S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
$\Rightarrow$ sample space of event is given by n(S) = 12
Let A be the event having sum of numbers is 3
$\therefore$ A = {(1, 2)}
$\Rightarrow$ n (A) = 1
Thus, P(A) = $\frac{1}{{12}}$
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Question 82 Marks
Given $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$. Find P (A or B), if A and B are mutually exclusive events.
Answer
Here $P(A) = \frac{3}{5}\ , P(B) = \frac{1}{5}$
Since A and B are mutually exclusive events
$\therefore \;P(A \cup B) = P(A) + P(B)$
$\therefore \;P(A \cup B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$
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Question 92 Marks
Three coins are tossed once. Let A denote the event three heads show, B denote the event two heads and one tail show, C denote the event three tails show and D denote the event a head shows on the first coin. Find Compound events?
Answer
Given that three coins are tossed. So sample space (S) is given by

S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Event A: three heads show = {HHH}
Event B: two heads and one tail show = {HHT, HTH, THH}
Event C: three tails show = {TTT}
Event D: a head shows on the first coin - {HHH, HHT, HTH, HTT}
A compound event is the occurrence of two or more outcomes together.
We see that, B = {HHT, HTH, THH}, and D = {HHH, HHT, HTH, HTT}.
Since cardinal numbers of events B and D are respectively 3 and 4,
So, B and D are compound events.

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Question 102 Marks
Three coins are tossed once. Let A denote the event three heads show, B denote the event two heads and one tail show, C denote the event three tails show and D denote the event a head shows on the first coin. Find Simple events?
Answer
Given that three coins are tossed then the sample space (S) is given by

S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Event A: three heads show = {HHH}
Event B: two heads and one tail show = {HHT, HTH, THH}
Event C: three tails show = {TTT}
Event D: a head shows on the first coin - {HHH, HHT, HTH, HTT}
Simple events have only one outcome
A ={HHH}, Here cardinal number = 1
$\therefore$ A is a simple event.
Since, C = {TTT}, Here also cardinal number = 1
$\therefore$ C is a simple event.

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Question 112 Marks
Three coins are tossed once. Let $A$ denote the event three heads show, $B$ denote the event two heads and one tail show, $C$ denote the event three tails show and $D$ denote the event a head shows on the first coin. find Mutually exclusive?
Answer
When three coins are tossed then the sample space $(S) =2^3 = 8$ and it is given by
$\text{S = \{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT\}}$
$A:$ three heads show $\text{= {HHH}}$
$B:$ two heads and one tail show $\text{= {HHT, HTH, THH}}$
$C:$ three tails show $\text{= {TTT}}$
$D:$ a head shows on the first coin $\text{- {HHH, HHT, HTH, HTT}}$
Here we have to examine which of the above events are mutually exclusive.
As we know that if two events do not have any common element
i.e., if intersection of two events is $\style{font-size:24px}\phi$ ,
then those events are called mutually exclusive events.
Here,clearly $\style{font-size:28px}{A\cap B=\phi}$, $\style{font-size:28px}{B\cap C=\phi}$, $\style{font-size:28px}{A\cap C=\phi}$ , $\style{font-size:28px}{C\cap D=\phi}$
Hence, events $A$ and $B,$ events $B$ and $C$, events $A$ and $C,$ events $C$ and $D$ are mutually exclusive events.
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Question 122 Marks
A coin is tossed three times, consider the events"
A: ‘No head appears’,
B: ‘Exactly one head appears’ and
C: ‘Atleast two heads appear’.
Do they form a set of mutually exclusive and exhaustive events?
Answer
Given that three coins are tossed. The sample space of the experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Now, outcomes of event A = {TTT},
Outcomes of event B = {HTT, THT, TTH},
Outcomes of event C = {HHT, HTH, THH, HHH}
Now
Since A $\cup$ B $\cup$ C = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S
Therefore, A, B and C are exhaustive events.
Also, A $\cap$ B = $\phi$, A $\cap$ C = $\phi$ and B $\cap$ C = $\phi$
Therefore, the events are pair-wise disjoint, i.e., they are mutually exclusive.
Hence, A, B and C form a set of mutually exclusive and exhaustive events.
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Question 132 Marks
Two dice are thrown and the sum of the numbers which come upon the dice is noted. Let us consider the following events associated with this experiment
A: the sum is even.
B: the sum is a multiple of 3.
C: the sum is less than 4.
D: the sum is greater than 11.
Which pairs of these events are mutually exclusive?
Answer
Two dice are thrown so there are 36 elements in the sample space S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}.
Then
A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
B = {(1, 2), (2, 1), (1, 5), (5, 1), (3, 3), (2, 4), (4, 2), (3, 6), (6, 3), (4, 5), (5, 4), (6, 6)}
C = {(1, 1), (2, 1), (1, 2)} and
D = {(6, 6)}
For mutually exclusive events, there should be no element common,
We find that A $\cap$ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)} $\ne$ $\phi$
Therefore, A and B are not mutually exclusive events.
Similarly A $\cap$ C $\ne$ $\phi$, A $\cap$ D $\ne$ $\phi$, B $\cap$ C $\ne$ $\phi$ and B $\cap$ D $\ne$ $\phi$.
Thus, the pairs of events, (A, C), (A, D), (B, C), (B, D) are not mutually exclusive events.
Also C $\cap$ D = $\phi$ and so C and D are mutually exclusive events.
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Question 142 Marks
Find the sample space associated with the experiment of rolling a pair of dice (one is blue and the other red) once. Also, find the number of elements of this sample space.
Answer
Since in each die the number of outcomes are 6.
So, the number of elements of this sample space is 6 × 6 = 36
Each outcome can be denoted by the ordered pair (x, y), where x is the number appeared on the blue die and y is the number appeared on the red die.
Therefore, this sample space is given by
S = {(x, y): x is the number on the blue die and y is the number on the red die}.
The sample space is given below:
S= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
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