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Probability · MP Board (MPBSE) STD 11 Science (Madhya Pradesh - 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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1 Marks Question - MATHS STD 11 Science Questions - Vidyadip