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Maths 3 Marks Question

P-1 Probability · Maharashtra Board STD 10 (MSBSHSE - English Medium). 36 questions.

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3 Marks Question

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Question 13 Marks
A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,
(1) a red balloon
(2) a blue balloon
(3) a green balloon.
Answer
Probability that Pranali gets a red balloon, $p(R)=\frac{\text { Number of red balloons }}{\text { Total number of balloons }}$
$\therefore p ( V )=\frac{4}{9}$
(2) Probability that Pranali gets a blue balloon, $p(B)=\frac{\text { Number of blue balloons }}{\text { Total number of balloons }}$
$\therefore p(B)=\frac{1}{3}$
(3) Probability that Pranali gets a green balloon, $p(G)=\frac{\text { Number of green balloons }}{\text { Total number of balloons }}$
$\therefore p ( G )=\frac{4}{9}$
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Question 23 Marks
The faces of a die bear numbers 0, 1, 2, 3, 4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.
Answer
Case 1: The face of die have number 0 when rolled twice.
$\therefore$ Probability that the product of digits on the upper face is zero=$=\frac{1}{6} \times \frac{1}{6}$
$=\frac{1}{36}$
Case 2: One face of die has 0 and the other has any number from 1 to 5
$\therefore$ Probability that the product of digits on the upper face is zero= $=\frac{1}{6} \times \frac{5}{6}$
$=\frac{5}{36}$
$\therefore$ Total probability that the product of digits on the upper face is zero= $=\frac{1}{36} + \frac{5}{36}$
$=\frac{6}{36}$
$=\frac{1}{6}$
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Question 33 Marks
A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a -
(1) prime number
(2) multiple of 4
(3) multiple of 11.
Answer
Sample Space, S= {10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32,33, 34, 40, 41, 42, 43, 44}
Number of sample points, n(S) = 20
(1) Probability that the number so formed is a prime number , p(Prime)= 6/20 = 3/10
$\therefore$ p(Prime) = 3/10
(2) Probability that the number so formed is a multiple of 4, p(M) = 6/20 = 3/10
$\therefore$ p(M) = 3/10
(3) Probability that the number so formed is a multiple of 11, p(M) = 4/20 = 1/5
$\therefore$ p(M) = 1/5
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Question 43 Marks
Out of 200 students from a school, 135 like Kabbaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected dosen't like Kabbaddi.
Answer
Number of students who do not like Kabbaddi=200-135=65
$\text { Probability that the student selected dosen't like Kabbaddi, } p(K)=\frac{\text { Number of students who do not like Kabbaddi }}{\text { Total number of students }} $
$ \therefore p(K)=\frac{13}{40}$
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Question 53 Marks
Each card bears one letter from the word ‘mathematics’ The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.
Answer
Probability that a card drawn bears the letter ' $m{ }^{\prime}, p(m)=\frac{\text { Number of } / m \prime}{\text { Total number of alphabets }}$
$\therefore p ( m )=\frac{2}{11}$
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Question 63 Marks
A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is -
(1) red.
(2) not red
(3) either red or white.
Answer
Probability that ball drawn is red, $p(R)=\frac{\text { Number of red balls }}{\text { Total number of balls }}$
$\therefore p(R)=\frac{3}{9}=\frac{1}{3}$
(2) Probability that ball drawn is not red, $p(N)=\frac{\text { Number of balls which are not red }}{\text { Total number of balls }}$
$\therefore p ( N )=\frac{6}{9}=\frac{2}{3}$
(3) Probability that ball drawn is either red or white, $p(E)=\frac{\text { Number of balls which are either red or white }}{\text { Total number of balls }}$
$\therefore p(E)=\frac{6}{9}=\frac{2}{3}$
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Question 73 Marks
A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
(1) an odd number
(2) a complete square number.
Answer
Probability of an event that the ticket drawn bears an odd number, $p ( O )=\frac{\text { Number of tickets with odd number }}{\text { Total number of tickets }}$
$\therefore p ( O )=\frac{1}{2}$
(2) Probability of an event that the ticket drawn bears a complete square number, $p(S)=\frac{\text { Number of tickets with square number }}{\text { Total number of tickets }}$
$\therefore p(S)=\frac{5}{30}=\frac{5}{6}$
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Question 83 Marks
Six faces of a die are as shown below.

If the die is rolled once, find the probability of -
(1) ‘A’ appears on upper face.
(2) ‘D’ appears on upper face.
Answer
Probability that $A$ appears on upper face, $p(A)=\frac{\text { Number of faces with } A}{\text { Total number of faces }}$
$\therefore p ( A )=\frac{2}{6}=\frac{1}{3}$
(2) Probability that $D$ appears on upper face, $p(D)=\frac{\text { Number of faces with } D}{\text { Total number of faces }}$
$\therefore p(D)=\frac{1}{6}$
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Question 93 Marks
A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?
Answer
Probability that Rutuja picks blue pen, $p(B)=\frac{\text { Number of blue pen }}{\text { Total number of pens }}$
$\therefore p(B)=\frac{8}{16}=\frac{1}{2}$
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Question 103 Marks
Three coins are tossed together. Find the probability of getting exactly two heads.
Answer
$\frac{3}{8}$
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Question 113 Marks
A box contains 300 electrical bulbs out of which 18 are defective. What is the probability that bulb chosen will not be defective?
Answer
$\frac{141}{150}$
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Question 133 Marks
Tickets numbered 1 to 30 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket drawn will be a multiple of 7 ?
Answer
$\frac{2}{15}$
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Question 143 Marks
A box contains 3 apples, 4 oranges and 5 bananas. One fruit is drawn at random from the box. Write $S, n(S)$ and sample points of each of the following events.
Event A: Fruit is orange or banana
Event B: Fruit is not an apple.
Event C: Fruit is neither apple nor banana
Event D : Fruit is banana
Answer
$n(S)=12, n(A)=9, n(B)=9, n( C )=4, n( D )=5$
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Question 153 Marks
A card is picked up randomly from well shuffled pack of cards. Write the $n(S), n(A), n(B)$ and $n( C )$
Event A: A red face card.
Event B: An ace of spade
Event C: Not a black king
Answer
$n(S)=52, n(A)=6, n(B)=1, n( C )=50$
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Question 163 Marks
An urn contains $10$ red and $8$ white balls. Write sample space $S$ and $n(S)$. Write the events $A$ and $B$ using Set and mention $n(A)$ and $n(B)$ if $A$ is the event that ball is white, $B$ is the event that ball is neither white nor red.
Answer
$ S =\left\{ R _{ 1 ^{\prime}} R _{2^{\prime}} R _{3^{\prime}} R _{ 4 ^{\prime}} R _{5^{\prime}} R _{6^{\prime}} R _{7^{\prime}} R _{8^{\prime}} R _{9^{\prime}} R _{ 10 },W _{1^{\prime}} W _{2^{\prime}}\right.$
$\left. W _{3^{\prime}} W _{4^{\prime}} W _{5^{\prime}} W _{6^{\prime}} W _{7^{\prime}} W _8\right\} n(S)=18$
$A=\left\{ W _{1^{\prime}}, W _{2^{\prime}} W _{3^{\prime}} W _{4^{\prime}} W _{5^{\prime}} W _{6^{\prime}} W _{7^{\prime}} W _{ 8 }\right\}=n(A)=8$
$B=n(B)=0$
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Question 173 Marks
A box contains cards numbered from $1$ to $30$ . Write the sample space $S$ and no. of sample points $n(S)$ and if a card is drawn at random, write $A$ and $n(A)$ if the card is divisible by $5.$
Answer
$ S =\{1,2,3,4,5, \ldots \ldots 30\} n(S)=30$
$A=\{5,10,15,20,25,30\} n(A)=6$
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Question 183 Marks
A card is drawn at random from a well shuffled pack of 52 cards. Find the probability that the card drawn is
(i) neither ace nor king.
(ii) black and a king
(iii) 10 of spades.
Answer
(i) $\frac{11}{13}$ (ii) $\frac{1}{26}$ (iii) $\frac{1}{52}$
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Question 193 Marks
A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
(1) an odd number
(2) a complete square number.

Answer
Probability of an event that the ticket drawn bears an odd number, $p(O)=\frac{\text { Number of tickets with odd number }}{\text { Total number of tickets }}$
$\therefore p(O)=\frac{1}{2}$
(2) Probability of an event that the ticket drawn bears a complete square number, $p(S)=\frac{\text { Number of tickets with square number }}{\text { Total number of tickets }}$
$\therefore p(S)=\frac{5}{30}=\frac{1}{6}$
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Question 203 Marks
Answer
Probability that $A$ appears on upper face, $p(A)=$ $p(A)=\frac{\text { Number of faces with A}}{\text { Total number of faces }}$
$\therefore p(A)=\frac{2}{6}=\frac{1}{3}$
(2) Probability that $D$ appears on upper face, $p(D)=\frac{\text { Number of faces with D}}{\text { Total number of faces }}$
$\therefore p(D)=\frac{1}{6}$
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Question 213 Marks
A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?

Answer
Probability that Rutuja picks blue pen, $p(B)=\frac{\text { Number of blue pen }}{\text { Total number of pens }}$
$\therefore p(B)=\frac{8}{16}=\frac{1}{2}$
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Question 223 Marks
A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,
(1) a red balloon
(2) a blue balloon
(3) a green balloon.

Answer
Number of red balloons
Probability that Pranali gets a red balloon, $p(R)=\frac{\text{Number of red balloons }}{\text{Total number of balloons }}$
$\therefore p ( V )=\frac{4}{9}$
Number of blue balloons
(2) Probability that Pranali gets a blue balloon, $p(B)=\frac{\text{Number of blue balloons }}{\text{Total number of balloons }}$
$\therefore p(B)=\frac{1}{3}$
Number of green balloons
(3) Probability that Pranali gets a green balloon, $p(G)=$ $\frac{\text{Number of green balloons }}{\text{Total number of balloons }}$
$\therefore p ( G )=\frac{4}{9}$
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Question 233 Marks
The faces of a die bear numbers 0, 1, 2, 3, 4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.

Answer
Case 1: The face of die have number 0 when rolled twice.
$\therefore$ Probability that the product of digits on the upper face is zero $=\frac{1}{6} \times \frac{1}{6}$
$=\frac{1}{36}$
Case 2: One face of die has 0 and the other has any number from 1 to 5
$\therefore$ Probability that the product of digits on the upper face is zero $=\frac{1}{6} \times \frac{5}{6}$
$=\frac{5}{36}$
$\therefore$ Total probability that the product of digits on the upper face is zero $=\frac{1}{36}+\frac{5}{36}$
$\begin{array}{l}
=\frac{6}{36} \\
=\frac{1}{6}
\end{array}$
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Question 243 Marks
A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a -
(1) prime number
(2) multiple of 4
(3) multiple of 11.
Answer
Sample Space,
 $S=\{10,11,12,13,14,20,21,22,23,24,30,31,32,33,34,40,41,42,43,44\}$
Number of sample points, $n ( S )=20$
(1) Probability that the number so formed is a prime number, $p($ Prime $)=6 / 20=3 / 10$
$\therefore p(\text { Prime })=3 / 10$
(2) Probability that the number so formed is a multiple of $4, p(M)=6 / 20=3 / 10$
$\therefore p(M)=3 / 10$
(3) Probability that the number so formed is a multiple of $11, p(M)=4 / 20=1 / 5$
$\therefore p(M)=1 / 5$
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Question 253 Marks
Out of 200 students from a school, 135 like Kabbaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected dosen't like Kabbaddi.

Answer
Number of students who do not like Kabbaddi=200-135=65
$\begin{array}{l}
\text { Probability that the student selected dosen't like Kabbaddi, } p(K)=\frac{\text { Number of students who do not like Kabbaddi }}{\text { Total number of students }} \\
\therefore p(K)=\frac{13}{40}
\end{array}$
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Question 263 Marks
Each card bears one letter from the word ‘mathematics’ The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.

Answer
Probability that a card drawn bears the letter 'm', $p(m)=\frac{\text { Number of } ' m '}{\text { Total number of alphabets }}$
$\therefore p(m)=\frac{2}{11}$
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Question 273 Marks
A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is -
(1) red.
(2) not red
(3) either red or white.

Answer
Probability that ball drawn is red, $p(R)=\frac{\text { Number of red balls }}{\text { Total number of balls }}$
$\therefore p(R)=\frac{3}{9}=\frac{1}{3}$
(2) Probability that ball drawn is not red, $p(N)=\frac{\text { Number of balls which are not red }}{\text { Total number of balls }}$
$\therefore p(N)=\frac{6}{9}=\frac{2}{3}$
(3) Probability that ball drawn is either red or white, $p(E)=\frac{\text { Number of balls which are either red or white }}{\text { Total number of balls }}$
$\therefore p(E)=\frac{6}{9}=\frac{2}{3}$
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Question 283 Marks
Tickets numbered 1 to 30 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket drawn will be a multiple of 7 ?
Answer
$\frac{2}{15}$
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Question 293 Marks
Three coins are tossed together. Find the probability of getting exactly two heads.
Answer
$\frac{3}{8}$
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Question 313 Marks
An urn contains 10 red and 8 white balls. Write sample space $S$ and $n(S)$. Write the events $A$ and B using Set and mention $n(A)$ and $n(B)$ if $A$ is the event that ball is white, B is the event that ball is neither white nor red.
Answer

$\begin{array}{l} S =\left\{ R _{ 1 ^{\prime}} R _{2^{\prime}} R _{3^{\prime}} R _{ 4 ^{\prime}} R _{5^{\prime}} R _{6^{\prime}} R _{7^{\prime}} R _{8^{\prime}} R _{9^{\prime}} R _{ 10 },W _{1^{\prime}} W _{2^{\prime}}\right. \\ \left. W _{3^{\prime}} W _{4^{\prime}} W _{5^{\prime}} W _{6^{\prime}} W _{7^{\prime}} W _8\right\} n(S)=18 \\ A=\left\{ W _{1^{\prime}}, W _{2^{\prime}} W _{3^{\prime}} W _{4^{\prime}} W _{5^{\prime}} W _{6^{\prime}} W _{7^{\prime}} W _{ 8 }\right\}=n(A)=8 \\ B=\{\quad\}=n(B)=0\end{array}$
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Question 323 Marks
A card is picked up randomly from well shuffled pack of cards. Write the $n(S), n(A), n(B)$ and $n( C )$
Event A: A red face card.
Event B: An ace of spade
Event C: Not a black king
Answer
$n(S)=52, n(A)=6, n(B)=1, n( C )=50$
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Question 333 Marks
A card is drawn at random from a well shuffled pack of 52 cards. Find the probability that the card drawn is
(i) neither ace nor king.
(ii) black and a king
(iii) 10 of spades.
Answer
(i) $\frac{11}{13}$ (ii) $\frac{1}{26}$ (iii) $\frac{1}{52}$
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Question 343 Marks
A box contains cards numbered from 1 to 30 . Write the sample space $S$ and no. of sample points $n(S)$ and if a card is drawn at random, write $A$ and $n(A)$ if the card is divisible by 5.
Answer

$\begin{array}{l} S =\{1,2,3,4,5, \ldots \ldots 30\} n(S)=30 \\ A=\{5,10,15,20,25,30\} n(A)=6\end{array}$
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Question 353 Marks
A box contains 3 apples, 4 oranges and 5 bananas. One fruit is drawn at random from the box. Write $S, n(S)$ and sample points of each of the following events.
Event A: Fruit is orange or banana
Event B: Fruit is not an apple.
Event C: Fruit is neither apple nor banana
Event D : Fruit is banana
Answer
$n(S)=12, n(A)=9, n(B)=9, n( C )=4, n( D )=5$
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Question 363 Marks
A box contains 300 electrical bulbs out of which 18 are defective. What is the probability that bulb chosen will not be defective?
Answer
$\frac{141}{150}$
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