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Probability · CBSE STD 11 Science (CBSE - English Medium). 85 questions.

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Question 11 Mark
If 4-digit numbers greater than 5000 are randomly formed from the digits 0, 1, 3, 5, and 7. What is the probability of forming a number divisible by 5 when the repetition of digits is not allowed?
Answer
We have to find the probability of forming a number divisible by 5 when the repetition of digits is not allowed
When digits are not repeated: In a 4-digit number greater than 5000, the thousandth place can be filled up by either 5 or 7.
If the thousandth place is filled by 5 then the other three places can be filled in = {4 $\times$ 3 $\times$ 2 = 24} ways.
Similarly when the thousandth place is filled by 7 then the other three places can be filled in = 4 $\times$ 3 $\times$ 2 = 24 ways.
$\therefore$ Without repeating digits, total number of 4-digit numbers greater than 5000 can be formed = 24 + 24 = 48.
Now to find the number of 4-digit numbers greater than 5000 and divisible by 5 (without repetition ).
A number greater than 5000 and divisible by 5 when the unit place is either 0 or 5 and the thousandth place is either 5 or 7.
case-I: When thousandth place is filled by 5, then the unit place will be filled by 0 (zero)
and a number of such numbers = 3 $\times$ 2 = 6.
case-II: When thousandth place is filled by 7, then unit place will be filled by either 0 (zero) or 5
and number of such numbers = 2(3 $\times$ 2) = 12.
$\therefore$ Without repetition, the total number of 4-digit numbers greater than 5000
and divisible by 5 = 6 + 12 = 18. [by case-I and case-II]
Hence, the required probability $={\frac{18}{48}=\frac38}$
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Question 21 Mark
If $4-$digit numbers greater than 5000 are randomly formed from the digits $0, 1, 3, 5,$ and $7$ what is the probability of forming a number divisible by $5$ when the digits are repeated?
Answer
We have to find the probability of forming a number divisible by $5$ when the digits are repeated
When digits are repeated:
In a $4-$digit number greater than $5000,$ the thousandth place can be filled up by either $5$ or $7.$
So the thousandth place can be filled in $2$ ways.
Since the digits can be repeated, so the remaining three places can be filled in $5^3= 125$ ways.
$\therefore$ Total number of $4$ digit numbers greater than $5000 = 2 \times 125 - 1 = 250 - 1 = 249.$
A number is divisible by $5$ if the digit at the unit place is either 0 or 5
For a $4$-digit number greater than $5000$ and divisible by $5$ the unit and thousandth place are fixed in $\{2 \times 2 = 4\}$ ways.
The hundredth and tenth place can be filled in $5^2 = 25$ ways.
Since the numbers are greater than $5000,$ so the number of $4-$digit numbers divisible by $5= 2\times5\times5\times2-1=100-1=99$
$\therefore$ The required probability of forming a $4-$digit number greater than $5000$
and divisible by $5$ when digits can be repeated = ${\frac{99}{249}=\frac{33}{83}}$
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Question 31 Mark
In a certain lottery 10,000 tickets are sold and, ten equal prizes are awarded. What is the probability of not getting a prize if you buy 10 tickets?
Answer
We have to find the probability of not getting a prize if you buy 10 tickets
Given, the total number of tickets = 10,000
So out of 10,000 tickets, one can buy 1 ticket in ${}^{10000}C_1=10,000\;ways$
Given, number of prize bearing tickets = 10
So number of tickets not bearing prize = 10,000 - 10 = 9990
Let C be the event that ten bought tickets are not prize bearing tickets.
Now, out of 10,000 tickets, one can buy 2 tickets in ${}^{10000}C_{10}$ ways
and out of 9990 tickets not bearing any prize, one can buy 10 tickets in ${}^{9990}C_{10}$ ways.
$\therefore$ $P(C)=\frac{{}^{9990}C_{10}}{{}^{10000}C_{10}}$
Hence, the probability of not getting a prize if ten tickets are bought $=\frac{{}^{9990}C_{10}}{{}^{10000}C_{10}}$.
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Question 41 Mark
In a certain lottery 10,000 tickets are sold and, ten equal prizes are awarded. What is the probability of not getting a prize if you buy two tickets?
Answer
Given, total number of tickets = 10,000
Number of prize bearing tickets = 10
So number of tickets not bearing prize = 10,000-10 = 9990
Let B be the event that two bought tickets are not prize bearing tickets.
Now, out of 10,000 tickets one can buy 2 ticket in ${}^{10000}C_2$ ways i.e. total number of outcomes,
and out of 9990 tickets not bearing any prize, one can buy 2 tickets in ${}^{9990}C_2$ways i.e. total number of favourable outcomes.
$\therefore$$P(B)=\frac{Number of favourable outcomes}{Total number of outcomes}=\frac{{}^{9990}C_2}{{}^{10000}C_2}$
Hence, probability of not getting a prize if two tickets are bought $=\frac{{}^{9990}C_2}{{}^{10000}C_2}$
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Question 51 Mark
In a certain lottery 10,000 tickets are sold and, ten equal prizes are awarded. What is the probability of not getting a prize if you buy one ticket?
Answer
Given, total number of tickets = total number of outcomes = 10,000
Also given, number of prize bearing tickets = 10
So number of tickets not bearing prize = 10,000-10 = 9990
Let A represent the event that one bought ticket is not bearing prize.
$\therefore$ $P(A)=\frac{9990}{10000} =\frac{999}{1000}$
Hence, probability of not getting a prize if one ticket is bought $=\frac{999}{1000}$
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Question 61 Mark
A die has two faces each with number 1, three faces each with number 2 and one face with number 3. If die is rolled once, determine P(not 3)
Answer
Total number of faces in a die = 6
Number of faces with number 3 = 1
$\therefore\;P(3) = \frac{1}{6}$
P (not 3) = 1 - P(3) $ = 1 - \frac{1}{6} = \frac{5}{6}$
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Question 71 Mark
A die has two faces each with number 1, three faces each with number 2 and one face with number 3. If die is rolled once, determine P(1 or 3)
Answer
Total number of faces in a die = 6
Number of faces with number 1 = 2
Number of faces with number 3 = 1
$\therefore \;P(1) = \frac{2}{6} = \frac{1}{3},\;P(3) = \frac{1}{6}$
P (1 or 3) = P(1) + P(3) $= \frac{1}{3} + \frac{1}{6} = \frac{{2 + 1}}{6} = \frac{3}{6} = \frac{1}{2}$
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Question 81 Mark
A die has two faces each with number 1, three faces each with number 2 and one face with number 3. If die is rolled once, determine P(2)
Answer
Total number of faces in a die = 6
Number of faces with number 2 = 3
$\therefore \;P(2) = \frac{3}{6}=\frac12$
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Question 91 Mark
Three coins are tossed once. Find the probability of getting: almost two tails
Answer
When three coins are tossed then the sample space S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting at most 2 tails
n(A) = 7
P(getting almost 2 tails) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{7}{\mathrm{8}}$
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Question 101 Mark
Three coins are tossed once. Find the probability of getting: no tail
Answer
When three coins are tossed then Sample space S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
And thus here n(S) = 8
Let A be the event of getting no tails
n(A) = 1
P(getting no tails) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{1}{\mathrm{8}}$
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Question 111 Mark
Three coins are tossed once. Find the probability of getting: exactly two tails
Answer
When three coins are tossed then total outcomes S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting exactly 2 tails
n(A) = 3
P(getting exactly 2 tails) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3}{\mathrm{8}}$
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Question 121 Mark
Three coins are tossed once. Find the probability of getting: 3 tails
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting 3 tails
n(A) = 1
P(getting 3 tails) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{1}{\mathrm{8}}$
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Question 131 Mark
Three coins are tossed once. Find the probability of getting: no head
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting no heads
n(A) = 1
P(getting no head) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{1}{\mathrm{8}}$
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Question 141 Mark
Three coins are tossed once. Find the probability of getting: at most 2 heads
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
let A be the event of getting at most 2 heads
n(A) = 7
P(getting at most 2 heads) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{7}{8}$
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Question 151 Mark
Three coins are tossed once. Find the probability of getting: at least 2 heads
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting at least 2 head
n(A) = 4
P(getting atleast 2 heads) = P(A) = $\frac{n(A)}{n(S)}=\frac{4}{8}=\frac{1}{2}$
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Question 161 Mark
Three coins are tossed once. Find the probability of getting: 2 heads
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting 2 heads
n(A) = 3
P(getting 2 heads) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3}{\mathrm{8}}$
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Question 171 Mark
Three coins are tossed once. Find the probability of getting: 3 heads
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting 3 heads n(A) = 1
P(getting 3 heads) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{1}{\mathrm{8}}$
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Question 181 Mark
A card is selected from a pack of 52 cards. Calculate the probability that the card is black card.
Answer
Let C be the event of drawing a black card. Now, we know that in a pack of 52 cards, there are 26 black cards.
$\therefore P(C) =\frac{{number of favourable outcomes}}{{number of total outcomes}}= \frac{{26}}{{52}} = \frac{1}{2}$
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Question 191 Mark
A card is selected from a pack of 52 cards. Calculate the probability that the card is an ace.
Answer
Let B be the event of drawing an ace. We know that in a pack of 52 card there are four aces.
$\therefore P(B) = \frac{4}{{52}} = \frac{1}{{13}}$
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Question 201 Mark
A card is selected from a pack of 52 cards. Calculate the probability that the card is an ace of spades.
Answer
Let A be the event of drawing an ace of spades. Now there is only once ace of spade.
Therefore,
$ P(A) = \frac{1}{{52}}$
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Question 211 Mark
A card is selected from a pack of 52 cards. How many points are there in the sample space?
Answer
We have to draw One card from a pack of 52 cards.
$\Rightarrow$ Number of points in the sample space S = n(S) = 52.
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Question 221 Mark
A die is thrown, find the probability of A number less than 6 will appear.
Answer
Here the sample space of the event is given by
S = {1, 2, 3, 4, 5, 6}
$\therefore$ n(S) = 6
Let E be the event of getting a number less than 6
E = {1, 2, 3, 4, 5} $\Rightarrow$ n(E) = 5
Thus, $P(E) = \frac{{n(E)}}{{n(S)}} = \frac{5}{6}$
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Question 231 Mark
A die is thrown. Find the probability of getting a number more than 6.
Answer
We have to find the probability of getting a number of more than 6
The sample space associated with the random experiment of rolling a die is given by
S = {1, 2, 3, 4 ,5 , 6}. Clearly, there are 6 elements in S.
$\therefore$The total number of elementary events = 6.
Since no face of the die is marked with a number greater than 6.
So, favourable number of elementary events = 0
Hence, required probability = $\frac{0}{6}$= 0
In fact, the given event is an impossible event. So, the probability of its occurrence is zero.
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Question 241 Mark
A die is thrown, find the probability of A number less than or equal to 1 will appear.
Answer
In a throw of a die, sample space S = {1, 2, 3, 4, 5, 6}
$\therefore$ n(S) = 6
Let C be the event of getting a number less than or equal to 1
Elements of C = {1} $\Rightarrow$ n(C) = 1
Thus $P(C)=\frac{{n(C)}}{{n(S)}} = \frac{1}{6}$
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Question 251 Mark
A die is thrown. Find the probability of getting a number greater than or equal to 3.
Answer
We have to find the probability of getting a number greater than or equal to 3.
The sample space associated with the random experiment of rolling a die is given by
S = {1, 2, 3, 4 ,5 , 6}. Clearly, there are 6 elements in S.
$\therefore$ The total number of elementary events = 6.
A number greater than or equal to 3 is obtained, if we get any one of 3, 4,5 , 6 as an outcome.
So, favourable number of elementary events = 4
Hence, the required probability =$\frac{4}{6}=\frac{2}{3}$
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Question 261 Mark
A die is thrown. Find the probability that a prime number will appear.
Answer
As 2, 3, 5 are prime numbers up to 6, so the desired outcomes are 2, 3, 5, and total outcomes are 1, 2, 3, 4, 5, 6
Therefore, total no. of outcomes are 6, and total no. of desired outcomes are 3
Probability of getting a prime number = $\frac{3}{6}$ = $\frac{1}{2}$
Conclusion: Probability of getting a prime number when a die is thrown is $\frac{1}{2}$
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Question 271 Mark
A coin is tossed twice, what is the probability that atleast one tail occurs?
Answer
Since a coin tossed twice,
so the sample space (S) is given by S= {HH, HT, TH, TT}
$\therefore$ Total number of possible out comes n (S) = 4
Let E be the event of getting at least one tail
$\therefore$ n(E) = 3
$\therefore$ Probability of getting at least one tail ${P(E)=\frac{n(E)}{n(S)}=\frac34}$
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Question 281 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = $\{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  $\frac{1}{14}$ $\frac{2}{14}$ $\frac{3}{14}$ $\frac{4}{14}$ $\frac{5}{14}$ $\frac{6}{14}$ $\frac{15}{14}$
Answer
The conditions of axiomatic approach in the given assignment are being fulfilled as $p(w_7)=\frac{15}{14}$>1 and probability should be less than or equal to one and positive.
Hence, the given assignment is not valid.
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Question 291 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = $\{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  $-0.1$ $0.2$ $0.3$ $0.4$ $-0.2$ $0.1$ $0.3$
Answer
The conditions of axiomatic approach do not hold true in the given assignment, because $p(w_1)$ and $p(w_4)$ is negative.
To be the conditions true each of the number $p(w_i)$ should be less than or equal to one and positive.
So, the assignment is not valid
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Question 321 Mark
P(A) P(B) P(A$\cap$B) P(A$\cup$B)
$\frac 13$ $\frac 15$ $\frac 1{15}$ _____
Answer
We know that
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
Putting Values
$P(A\cup B) = \frac 13 + \frac 15 - \frac 1{15}$
$P(A\cup B) = \frac {5\; +\; 3\; -\; 1}{15}$
$P(A\cup B) = \frac 7{15}$
Hence $P(A\cup B) = \frac 7{15}$
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Question 331 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  $0.1$ $0.2$ $0.3$ $0.4$ $0.5$ $0.6$ $0.7$
Answer
Both the conditions of axiomatic approach in the given assignment are
$i.$ Each of the number $p(w_i)$ is less than or equal to one and is positive,
$ii.$ Sum of probabilities is $0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1.$
It's clear that the second condition is not satisfied as the sum should be exactly equal to one.
Hence, the given assignment is not valid.
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Question 341 Mark
Check whether the probabilities P(A) and P(B) are consistently defined P(A) = 0.5, P(B) = 0.4, $P(A \cup B) = 0.8$
Answer
Given that P(A) = 0.5, P(B) = 0.4 and $P(A \cup B) = 0.8$
Applying the general addition rule,

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\therefore \;0.8 = 0.5 + 0.4 - P(A \cap B)$
$\Rightarrow \;P(A \cap B) = 0.9 - 0.8 = 0.1$
$\therefore \;P(A \cap B) < P(A)\;{\text{and}}\;P(A \cap B) < P(B)$
Thus the given probabilities are consistently defined.

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Question 351 Mark
In a simultaneous throw of a pair of dice, find the probability of getting an even number on one and a multiple of 3 on the other.
Answer
We have to find the probability of getting an even number on one and a multiple of 3 on the other.
Let E be the event of getting even on one and multiple of three on other.
E = {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}
n(E) = 11
$P(E)=\frac{n(E)}{n(S)}$
$P(E)=\frac{11}{36}$
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Question 361 Mark
Check whether the probabilities P(A) and P(B) are consistently defined P(A) = 0.5. P(B) = 0.7, $P(A \cap B) $ = 0.6
Answer
Given that P(A) = 0.5, P(B) = 0.7 and $P(A \cap B) = 0.6$
Now, as $P(A \cap B) > P(A)$,
We can say that the given probabilities are not consistently defined.
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Question 371 Mark
In a simultaneous throw of a pair of dice, find the probability of getting 8 as the sum.
Answer
Since a pair of dice have been thrown,
$\therefore$ Numbers of elementary events in sample space is $6^2 = 36$
Suppose E be the event that the sum 8 appear on the faces of dice,
$\therefore$ E = {(2,6), (3,5), (4,4), (5,3), (6,2)}
$\therefore$ $n(E) = 5$
$\therefore$ $P (E) =$ $\frac{5}{36}$
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Question 381 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$
Answer
Both the conditions of axiomatic approach hold true in the given assignment, that is
$i.$ Each of the number $p (w_i)$ is less than or equal to one and is positive
$ii.$ Sum of probabilities is $\frac 17 +\frac 17+\frac 17+\frac 17+\frac 17+\frac 17+\frac 17+\frac 77 = 1$
Hence, the given assignment is valid.
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Question 391 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  $0.1$ $0.01$ $0.05$ $0.03$ $0.01$ $0.2$ $0.6$
Answer
Both the conditions of axiomatic approach hold true in the given assignment, that is
  1. Each of the probability $p(w_i)$ is less than one and is positive
  2. Sum of probabilities is $0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1$
Hence,the given assignment is valid.
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Question 401 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event '$A \cap B'\cap C'$'
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (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)}
Event A: getting an even number on the first die.
Outcomes 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)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
B' = S - B
= {(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)} = A
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
C' = S - C
= {(1, 5), (1, 6),(2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6),(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)}
Clearly, $A \cap B' \cap C'$ = {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,6)}
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Question 411 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event B and C
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (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)}
Event A: getting an even number on the first die.
Outcomes 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)}
Event B: getting an odd number on the first die
Outcomes of event B ={(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
B and C = $B \cap C$ = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}
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Question 421 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\leq 5$
Describe the event 'B or C'
Answer
Given that two dice are thrown then space space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (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)}
Event A: getting an even number on the first die.
Outcomes 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)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\leq 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
B or C = $B \cup C$ =
{(1,1), (1,2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (5,1),(5,2), (5, 3), (5, 4), (5, 5), (5, 6)}
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Question 431 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event A but not C
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (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)}
Event A: getting an even number on the first die.
Outcomes 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)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
A but not C = A - C
= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
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Question 441 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event A and B
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (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)}
Event A: getting an even number on the first die.
Outcomes 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)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
A and B = $A \cap B = \phi$
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Question 451 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event A or B
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (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)}
Event A: getting an even number on the first die.
Outcomes 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)}
Event B: getting an odd number on the first die
Outcomes of event B ={(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

A or B =$A \cup B$ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, l), (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)} = S

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Question 461 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event not B
Answer
Given that two dice are thrown. So sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (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)}
Event A: getting an even number on the first die.
Outcomes 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)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C= {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
not B = {(2, l), (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)} = A
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Question 471 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event A'
Answer
Given that two dice are thrown, so sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (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)}
Event A: getting an even number on the first die.
Outcomes 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)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

A' = S - A
= (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} = B

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Question 481 Mark
Three coins are tossed. Describe Three events which are mutually exclusive but not exhaustive.
Answer
Given that three coins are tossed then sample space (S) is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let event A: getting three heads = {HHH},
event B: getting exactly two heads = {HHT, HTH, THH} and
event C: getting three tails = {TTT}
Now $\style{font-size:28px}{A\cap B=\phi}$, $\style{font-size:28px}{B\cap C=\phi}$ and $\style{font-size:28px}{A\cap C=\phi}$
Thus A, B, C are mutually exclusive events
Also ${A\cup B\cup C=\{HHH,\;HHT,\;HTH,\;THH,\;TTT\;\}⧧S}$
Thus A, B, C are mutually exclusive but not exhaustive events.

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Question 491 Mark
Three coins are tossed. Describe Two events which are mutually exclusive but not exhaustive.
Answer
Given that three coins are tossed then sample space (S) is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let event A: getting three heads = {HHH} and
event B: getting three tails = {TTT}
Now ${A\cap B=\phi}$ and ${AUB=\{HHH,\;TTT\;\}⧧S}$
Thus A and B are mutually exclusive but not exhaustive.
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Question 501 Mark
Three coins are tossed. Describe Two events, which are not mutually exclusive.
Answer
Given that three coins are tossed then sample space (S) is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let event A: getting at least two heads = {HHH, HHT, HTH, THH} and
event B: getting exactly two heads = {HHT, HTH, THH}
Now $\style{font-size:28px}{A\cap B=\{HHT,\;HTH,\;THH\}⧧\phi}$
$\therefore$ A and B are not mutually exclusive.
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Question 511 Mark
Three coins are tossed. Describe Three events which are mutually exclusive and exhaustive.
Answer
When three coins are tossed then sample space (S) is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Now, let A be the event: getting at least two heads = {HHH, HHT, HTH, THH}
B be the event: getting exactly one head = {HTT, THT, TTH}
and C be th event: getting no head = {TTT}

Mutually exclusive events are those in which no element is common
Since ${A\cap B=\phi}$. $\therefore$ A and B are mutually exclusive events,
$\style{font-size:28px}{B\cap C=\phi}$. $\therefore$ B and C are mutually exclusive events,
$\style{font-size:28px}{A\cap C=\phi}$. $\therefore$ A and C are mutually exclusive events.
Thus A, B, C are mutually exclusive events.
Also, as ${A\cup B\cup C=\{HHH,\;HHT,\;HTH,\;THH,\;HTT,\;THT,\;TTH,\;TTT\}=S}$
So, A, B, C are exhaustive events.
A, B, C are three events which are mutually exclusive and exhaustive.

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Question 521 Mark
Three coins are tossed. Describe Two events which are mutually exclusive.
Answer
When three coins are tossed then sample space (S) is given by S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let event A: getting at least two heads = {HHH, HHT, HTH, THH} and
event B: getting at least two tails = {HTT, THT, TTH, TTT}
There should not be any element common for the events to be mutually exclusive.
since ${A\cap B=\phi}$
Thus A and B are mutually exclusive events.
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Question 531 Mark
A die is rolled. Let E be the event die shows 4 and F be the event die shows even number. Are E and F mutually exclusive?
Answer
When a die is rolled then S = {1, 2, 3, 4, 5, 6}
E: die shows 4 = {4}
F: die shows even number = {2, 4.6}
Now $E \cap F = (4) \ne \phi$
Thus E and F are not mutually exclusive events.
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Question 541 Mark
Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}. Which of the following assignments of probabilities to each outcome are valid?
Outcomes $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$
0.1 0.2 0.3 0.4 0.5 0.6
Answer
Since, sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1.
Hence, the assignment is not valid.
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Question 551 Mark
Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}.Which of the following assignments of probabilities to each outcome are valid?
Outcomes $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$
$\frac 1{12}$ $\frac 1{12}$ $\frac 16$ $\frac 16$ $\frac 16$ $\frac 32$
Answer
We know that in no case P > 1.
Since p($\omega_6$) = $\frac32$> 1, the assignment is not valid.
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Question 561 Mark
Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}.Which of the following assignments of probabilities to each outcome are valid?
Outcomes $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$
$\frac 18$ $\frac 23$ $\frac 13$ $\frac 13$ $-\frac 14$ $-\frac 13$
Answer
Clearly we see that two of the probabilities p($\omega_5$) and p($\omega_6$) are negative so the assignment is not valid.
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Question 571 Mark
Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}.Which of the following assignments of probabilities to each outcome are valid?
Outcomes $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$
1 0 0 0 0 0
Answer
As we see that the given assignment is following both the conditions of axiomatic approach of probability
Condition (i): Each of the number p($\omega_i$) is 0 ≤p($\omega_i$)≤ 1
Condition (ii) Sum of the probabilities = 1 + 0 + 0 + 0 + 0 + 0 = 1
Therefore, the assignment is valid.
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Question 581 Mark
Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}.Which of the following assignments of probabilities to each outcome are valid?
Outcomes $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$
$\frac 16$ $\frac 16$ $\frac 16$ $\frac 16$ $\frac 16$ $\frac 16$
Answer
Given assignment is following both the conditions to be valid.
Condition (i): Each of the number p($\omega_i$) is positive and less than one.
Condition (ii): Sum of probabilities
$= \frac16 + \frac16 + \frac16+ \frac16+ \frac16+\frac16 = 1$
Therefore, the assignment is valid
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Question 591 Mark
Consider the experiment of rolling a die. Let A be the event getting a prime number, B be the event getting an odd number. Write the sets representing the event not A.
Answer
Here the sample space S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
‘not A’ = A′ = {1,4,6}
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Question 601 Mark
Consider the experiment of rolling a die. Let A be the event getting a prime number, B be the event getting an odd number. Write the sets representing the event A but not B.
Answer
Here sample space S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
‘A but not B’ = A – B = {2}
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Question 611 Mark
Consider the experiment of rolling a die. Let A be the event getting a prime number, B be the event getting an odd number. Write the sets representing the event A and B.
Answer
Here sample space S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
‘A and B’ = A $\cap$ B = {3,5}
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Question 621 Mark
Consider the experiment of rolling a die. Let A be the event getting a prime number, B be the event getting an odd number. Write the sets representing the event A or B
Answer
Here sample space s= {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
‘A or B’ = A $\cup$ B = {1, 2, 3, 5}
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Question 631 Mark
Consider the experiment in which a coin is tossed repeatedly until a head comes up. Describe the sample space.
Answer
The experiment will continue until we get a head.
So, head may come up on the first toss and the outcome will be H,
If first toss is tail and 2nd toss is head outcome will be TH,
In the same way, if first and second toss is tails and third toss is a head, outcome will be TTH and so on till head is obtained.
Hence, the desired sample space is S= {H, TH, TTH, TTTH, TTTTH,...}
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Question 641 Mark
A coin is tossed. If it shows head, we draw a ball from a bag consisting of $3$ blue and $4$ white balls; if it shows tail we throw a die. Describe the sample space of this experiment.
Answer
Let us denote three blue balls by $B_1, B_2, B_3 $nand four white balls by $W_1, W_2, W_3, W_4.$
Then a sample space of the experiment is
$S = \{ HB_1, HB_2, HB_3, HW_1, HW_2, HW_3, HW_4, T1, T2, T3, T4, T5, T6\}.$
Here $HB1$ means head on the coin and blue ball B1 is drawn, HW1 means head on the coin and white ball $W1$ is drawn. Similarly, $T1$ means tail on the coin and the number $1$ on the die.
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Question 651 Mark
Specify appropriate sample space. A person is noting down the number of accidents along a busy highway during a year.
Answer
The number of accidents along a busy highway during the year of observation can be either 0 (for no accident) or 1 or 2, or some other positive integer.
Thus, a sample space of the event is given by
S = {0, 1, 2, ...}
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Question 661 Mark
Specify appropriate sample space. A boy has a 1 rupee coin, a 2 rupee coin and a 5 rupee coin in his pocket. He takes out two coins out of his pocket, one after the other.
Answer
Let P denotes a 1 rupee coin, Q denotes a 2 rupee coin and R denotes a 5 rupee coin.
The first coin he takes out of his pocket may be any one of the three coins P, Q or R.
Corresponding to P, the second draw may be Q or R. So the result of two draws may be PQ or PR.
Similarly, corresponding to Q, the second draw may be R or P.
Therefore, the outcomes may be QP or QR.
Lastly, corresponding to R, the second draw may be P or Q.
So, the outcomes may be RP or RQ.
Thus, after combining all of the above outcomes, the sample space is S = {PQ, PR, QR, QP, RP, RQ}
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Question 671 Mark
In a relay race there are five teams $A, B, C, D$ and $E.$ What is the probability that $A, B$ and $C$ are first three to finish $($in any order$) ($Assume that all finishing orders are equally likely$)$
Answer
If we consider the sample space consisting of all finishing orders in the first three places, we will have $^5P_3, = \frac {5!}{(5-3)!}$ = $5 \times 4 \times 3$ = 60 = total number of outcomes
$A, B$ and $C$ are the first three finishers. There will be 3! arrangements for $A, B$ and $C.$ Therefore, the sample points corresponding to this event will be $3! = 6 =$ total number of favourable outcomes.
So, $P(A, B$ and $C$ are first three to finish$) = \frac{Total ~number~ of ~outcomes}{Total ~number~ of favourable~ outcomes}$$\frac {3!}{60} = \frac {6}{60} = \frac {1}{10}$
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Question 681 Mark
In a relay race there are five teams $A, B, C, D$ and $E.$ What is the probability that $A, B$ and $C$ finish first, second and third, respectively.
Answer
We have to find the the probability that $A, B$ and $C$ finish first, second and third, respectively.
If we consider the sample space consisting of all finishing orders in the first three places, we will have $^5P_3$,
i.e., $\frac {5!}{(5-3)!}  =  5 \times 4 \times 3 = 60$ sample points, each with a probability of $\frac 1{60}$.
A, B and C finish first, second and third, respectively. There is only one finishing order for this, i.e., $ABC$.
Thus $P(A, B$ and $C$ finish first, second and third respectively$) = \frac 1{60}$.
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Question 691 Mark
On her vacations, Veena visits four cities (A, B, C, and D) in random order. What is the probability of A just before B?
Answer
We have to find the value the probability of A just before B
The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e., 24.Therefore, n (S) = 24.
Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BDAC, BDCA, BCAD, BCDA,
CABD, CADB, CBDA, CBAD, CDAB, CDBA,
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G be the event ‘Veena visits A either first or second’.
Here G = {ABCD, ABDC, CABD, CDAB, DABC, DCAB}
Therefore, P(G) = $\frac {n(G)}{n(S)} = \frac {6}{24} = \frac 1{4}$
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Question 701 Mark
On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability of A either first or second?
Answer
The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e. 24.
Therefore, n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G be the event ‘Veena visits A either first or second’.
Here G = {ABCD, ABDC, ADBC, ACDB, ACBD, ADCB
BACD, BADC, CABD, CADB, DABC, DACB}
n(G) = 12
Therefore, P(G) = $\frac {n(G)}{n(S)} = \frac {12}{24} = \frac 1{2}$
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Question 711 Mark
On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability of A first and B last?
Answer
The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e. 24.
Therefore, total number of elements n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G be the event ‘Veena visits A first and B Last’.
Here G = {ACDB, ADCB}
n(G) = 2
Therefore, P(G) = $\frac {n(G)}{n(S)} = \frac 2{24} = \frac 1{12}$
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Question 721 Mark
On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability of A before B and B before C?
Answer
The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e. 24.
Therefore, total number of outcomes n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let the event ‘Veena visits A before B and B before C’ be denoted by F.
Here F = {ABCD, DABC, ABDC, ADBC}
n(F) = 4
Therefore, P(F) = $\frac {n(F)}{n(S)} = \frac 4{24} = \frac 16$
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Question 731 Mark
On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability of A before B?
Answer
The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e. 24.

Therefore, total number of outcomes, n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let the event ‘she visits A before B’ be denoted by E
Therefore, E = {ABCD, CABD, DABC, ABDC, CADB, DACB
ACBD, ACDB, ADBC, CDAB, DCAB, ADCB}
n(E) =12
Thus P(E) = $\frac {n(E)}{n(S)} =\frac {12}{24} = \frac 12$

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Question 741 Mark
A committee of two persons is selected from two men and two women. What is the probability that the committee will have two men?
Answer
It is given that, there are two men and two women.
Now, a committee of two persons is selected.
$\therefore$ $n (S) =\ ^4C_2 = \frac{4 \times 3}{2} = 6$
Let E be the event that two men in the committee
$\therefore n(E) =\ {^2C_2} = 1$
Therefore, $ P (E) =$ $\frac{1}{6}$
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Question 751 Mark
A committee of two persons is selected from two men and two women. What is the probability that the committee will have one man?
Answer
It is given that, there are two men and two women.
Now, a committee of two persons is selected.
$\therefore$ $n (S) =\ ^4C_2 =$ $\frac{4 \times 3}{2}$
Let E be the event that one man is in the committee
$\therefore$ $n (E) =\ ^2C_1 \times\ ^2C_1$
= 2 $\times $ 2 = 4
Therefore, P(E) = $\frac{4}{6}=\frac{2}{3}$
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Question 761 Mark
A committee of two persons is selected from two men and two women. What is the probability that the committee will have no man?
Answer
We have to find the probability that the committee will have no man
Given, there are two men and two women.
Now, a committee of two persons is selected.
$\therefore$ $n (S) =\ ^4C_2 =$ $\frac{4 \times 3}{2}$ = 6
Let E be the event that no man is to be in the committee
$\therefore$ $n (E) =\ ^2C_2 = 1$ [Only women will be in the committee]
$\therefore$ $P (E) = $$\frac{1}{6}$
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Question 771 Mark
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that only one of them will qualify the examination.
Answer
Let E and F be the events that Anil and Ashima will qualify the examination, respectively.
Given that
P(E) = 0.05, P(F) = 0.10 and P(E $\cap$ F) = 0.02.
The event only one of them will qualify the examination is same as the event either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima will qualify) i.e., E $\cap$ F´ or E´ $\cap$ F, where E $\cap$ F´ and E´ $\cap$ F are mutually exclusive
Therefore,
P(only one of them will qualify)
= P(E $\cap$ F´ or E´ $\cap$ F)
= P(E $\cap$ F´) + P(E´ $\cap$ F) - P [(E $\cap$ F´) $\cap$ P(E´ $\cap$ F)] [by general addition rule and also P [(E $\cap$ F´) $\cap$ P(E´ $\cap$ F)]= 0]
= P (E) – P(E $\cap$ F) + P(F) – P (E $\cap$ F)
= 0.05 – 0.02 + 0.10 – 0.02 = 0.11
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Question 781 Mark
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that at least one of them will not qualify the examination.
Answer
Let E and F denote the events that Anil and Ashima will qualify the examination, respectively.
Given that
P(E) = 0.05, P(F) = 0.10 and P(E $\cap$ F) = 0.02.
P(at least one of them will not qualify)
= 1 - P(both of them will qualify)
= 1 - P(E $\cap$ F)
= 1 - 0.02
= 0.98
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Question 791 Mark
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that 'Both Anil and Ashima will not qualify the examination.
Answer
Let E and F denote the events that Anil and Ashima will qualify the examination, respectively.
Given that
P(E) = 0.05, P(F) = 0.10 and P(E $\cap$ F) = 0.02.
Now,the event ‘both Anil and Ashima will not qualify the examination’ may be expressed as E´ $\cap$ F´.
Since, E´ is ‘not E’, i.e., Anil will not qualify the examination and F´ is ‘not F’, i.e., Ashima will not qualify the examination.
Also E´ $\cap$ F´ = (E $\cup$ F)´ (by Demorgan's Law)
So, P(E´ $\cap$ F´) = P(E $\cup$ F)´
Applying the formula,
P(E $\cup$ F) = P(E) + P(F) – P(E $\cup$ F)
or P(E ∪ F) = 0.05 + 0.10 – 0.02 = 0.13
Therefore P(E´ $\cap$ F´) = P(E $\cup$ F)´ = 1 – P(E $\cup$ F) = 1 – 0.13 = 0.87
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Question 801 Mark
One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be not a black card.
Answer
When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes = 52.
Let C be the event ‘card drawn is black card’
As we know that total number of black cards = 26
So P(C) = $\frac {26}{52} = \frac 12$
Thus, probability of a black card = $\frac 12$
The event ‘card drawn is not a black card’ may be denoted as C′ or ‘not C’.
We know that P(not C) = 1 – P(C) = $1 - \frac 12 = \frac 12$
Therefore, probability of not a black card = $\frac 12$
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Question 811 Mark
One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be not a diamond.
Answer
When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes = 52.
Let A be the event 'the card drawn is a diamond'
We know that total number of diamond cards = 13.
Therefore, P(A)$= \frac {13}{52} = \frac 14$
So the event ‘card drawn is not a diamond’ may be denoted as A' or ‘not A’
Now P(not A) = 1 – P(A) $= 1-\frac14 = \frac 34 $
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Question 821 Mark
One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be a black card (i.e., a club or, a spade).
Answer
When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52.
Let C be the event ‘card drawn is black card’.
Since total number of black cards = 26
So, P(C) = $\frac {26}{52} = \frac 12$
Thus, probability of a black card = $\frac 12$
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Question 831 Mark
One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be not an ace.
Answer
When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52.
Let A be the event ‘Card drawn is an ace’.
Since there are 4 cards of ace, so P(B) = $\frac{4}{52}$
We know that P($\bar B$) = 1 – P(B) = $= 1 - \frac 4{52} = 1 - \frac 1{13} = \frac {12}{13}$
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Question 841 Mark
One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be 'a diamond'
Answer
When a card is drawn from a well shuffled deck of 52 cards,
Total number of possible outcomes = 52.
Let A be the event 'the card drawn is a diamond'
Clearly the number of elements in set A is 13.
Total number of favorable outcomes = 13
Therefore, P(A) $= \frac {13}{52} = \frac 14$
i.e. probability of being a diamond card = $\frac 14$
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Question 851 Mark
Two coins (a one rupee coin and a two rupee coin) are tossed once. Find a sample space.
Answer
Since the coins are distinguishable so we can speak of the first coin and the second coin. Since either coin can turn up Head (H) or Tail (T), the possible outcomes may be
Heads on both the coins = (H,H) = HH
Head on first coin and Tail on the other = (H,T) = HT
Tail on first coin and Head on the other = (T,H) = TH
Tail on both coins = (T,T) = TT
Combination of all the above outcomes will give us the sample space
Thus, the sample space is S = {HH, HT, TH, TT}
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