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MATHS [1 Mark Questions]

Statistics and Probability · Tamil Nadu Board STD 10 (Tamil - English Medium). 32 questions.

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[1 Mark Questions]

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32 questions · timed · auto-graded

Question 11 Mark
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting a heart card bearing the number 5
Answer
Total number of cards $=52$
3 cards are removed
Remaining number of cards $=52-3=49$
$
n(S)=49
$
Let $D$ be the event of getting a 5 of heart card.
$
n(D)=1
$
$
P(D)=\frac{n(D)}{n(S)}=\frac{1}{49}
$
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Question 21 Mark
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting a diamond
Answer
Total number of cards $=52$
3 cards are removed

Remaining number of cards $=52-3=49$
$
n(S)=49
$

Let $A$ be the event of getting a diamond card.
$
\begin{aligned}
& n(A)=13 \\
& P(A)=\frac{n(A)}{n(S)}=\frac{13}{49}
\end{aligned}
$
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Question 31 Mark
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting a queen
Answer
Total number of cards $=52$
3 cards are removed

Remaining number of cards $=52-3=49$
$
n(S)=49
$

Let $B$ be the event of getting a queen card.
$
\begin{aligned}
& n(B)=(4-1)=3 \\
& P(B)=\frac{n(B)}{n(S)}=\frac{3}{49}
\end{aligned}
$
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Question 41 Mark
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting a spade
Answer
Total number of cards $=52$
3 cards are removed

Remaining number of cards $=52-3=49$
$n(S)=49$

Let $C$ be the event of getting a spade card.
$n(C)=(13-3)=10$
$P ( C )=\frac{ n ( C )}{ n ( S )}=\frac{10}{49}$
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Question 51 Mark
The probability that atleast one of $A$ and $B$ occur is 0.6 . If $A$ and $B$ occur simultaneously with probability 0.2 , then find $P (\overline{ A })+ P (\overline{ B })$
Answer
$\begin{aligned} & \text { Here } P(A \cup B)=0.6, P(A \cap B)=0.2 \\ & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & 0.6=P(A)+P(B)-0.2 \\ & P(A)+P(B)=0.8 \\ & P(\bar{A})+P(\bar{B})=1-P(A)+1-P(B) \\ & =2-[P(A)+P(B)] \\ & =2-0.8 \\ & =1.2\end{aligned}$
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Question 61 Mark
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A ∩ B) = 0.16. Find P(not A)
Answer
$\begin{aligned} & P(A)=0.42 \\ & P(B)=0.48 \\ & P(A \cap B)=0.16 \\ & P(\text { not } A)=P(\bar{A}) \\ & =1-P(A) \\ & =1-0.42 \\ & =0.58\end{aligned}$
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Question 71 Mark
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A ∩ B) = 0.16. Find P(not B)
Answer
$\begin{aligned} & P(A)=0.42 \\ & P(B)=0.48 \\ & P(A \cap B)=0.16 \\ & P(\text { not } B)=P(\bar{B})=1-P(B) \\ & =1-0.48 \\ & =0.52\end{aligned}$
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Question 81 Mark
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A ∩ B) = 0.16. Find P(A or B)
Answer
P(A) = 0.42
P(B) = 0.48
P(A ∩ B) = 0.16
P(A or B) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= 0.42 + 0.48 – 0.16
= 0.74
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Question 91 Mark
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is white
Answer
Sample space $(S)=5+6+7+8$
$n(S)=26$

Let $A$ be the event of getting a white ball
$
n(A)=6
$
$
P(A)=\frac{n(A)}{n(S)}
$
$
P(A)=\frac{6}{26}=\frac{3}{13}
$
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Question 101 Mark
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is black or red
Answer
Sample space $(S)=5+6+7+8$
$
n(S)=26
$

Let $A$ be the event of getting a black ball
$
n(A)=8
$
$
P(A)=\frac{n(A)}{n(S)}=\frac{8}{26}
$

Let $B$ be the event of getting a red ball
$
n(B)=5
$
$
P(B)=\frac{n(B)}{n(S)}=\frac{5}{26}
$

Probability of getting black or red ball
$
\begin{aligned}
& P(A \cup B)=P(A)+P(B) \\
& =\frac{8}{26}+\frac{5}{26} \\
& =\frac{13}{26}
\end{aligned}
$
$=\frac{1}{2}$
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Question 111 Mark
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is not white
Answer
Sample space $(S)=5+6+7+8$
$
n(S)=26
$

Let $A$ be the event of getting a white ball
$n(A)=6$
$
P(A)=\frac{n(A)}{n(B)}=\frac{6}{26}=\frac{3}{13} \cdots(1)
$

Not white probability of getting white ball
$P(A)=\frac{3}{13} \ldots \ldots . .($ from $(1))$

Probability of not getting white ball $P (\overline{ A })=1- P ( A )$
$
=1-\frac{3}{13}=\frac{13-3}{13}=\frac{10}{13}
$
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Question 121 Mark
Three fair coins are tossed together. Find the probability of getting atmost two tails
Answer
Three fair coins are tossed together
Sample spade $=\{H H H, H H T, H T H, H T T$, THH, THT, TTH, TTT $\}$
$n(S)=8$

Let $D$ be the event of getting atmost two tails.
$D=\{H T T, T T T, T T H, T H T, T H H, H H T, H T H\}$
$n(D)=7$
$P(D)=\frac{n(D)}{n(S)}=\frac{7}{8}$
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Question 131 Mark
Three fair coins are tossed together. Find the probability of getting all heads
Answer
Three fair coins are tossed together
Sample spade $=\{H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T\}$
$
n(S)=8
$

Let $A$ be the event of getting all heads
$
\begin{aligned}
& A=\{H H H\} \\
& n(A)=1 \\
& P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}
\end{aligned}
$
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Question 141 Mark
Three fair coins are tossed together. Find the probability of getting atleast one tail
Answer
Three fair coins are tossed together

Sample spade $=\{H H H, H H T, H T H, H T T$, THH, THT, TTH, TTT $\}$
$
n(S)=8
$

Let $B$ be the event of getting atleast one tail.
$B=\{H H T, H T H, H T T, T H H, T H T, T T H, T T T\}$
$
n(B)=7
$
$
P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}
$
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Question 151 Mark
Three fair coins are tossed together. Find the probability of getting atmost one head
Answer
$\begin{aligned} & \text { Three fair coins are tossed together } \\ & \text { Sample spade }=\{H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T\} \\ & n(S)=8 \\ & \text { Let } C \text { be the event of getting atmost one head } \\ & C=\{H T T, T H T, T T H, T T T\} \\ & n(C)=4 \\ & P(C)=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\end{aligned}$
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Question 161 Mark
Two unbiased dice are rolled once. Find the probability of getting the sum as 1
Answer
$
\begin{aligned}
& \text { Sample space }=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2) \text {, } \\
& (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) \text {, } \\
& (6,2),(6,3),(6,4),(6,5),(6,6)\} \\
& n(S)=36 \\
& n(D)=0 \\
& P(D)=\frac{n(D)}{n(S)}=\frac{0}{36}=0 \\
&
\end{aligned}
$
Let $D$ be the event of getting a sum is 1
Probability of getting a sum is 1 is 0
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Question 171 Mark
Two unbiased dice are rolled once. Find the probability of getting the product as a prime number
Answer
$
\begin{aligned}
& \text { Sample space }=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2), \\
& (3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1), \\
& (6,2),(6,3),(6,4),(6,5),(6,6)\} \\
& n(S)=36
\end{aligned}
$
Let $B$ be the event of getting a product is a prime number.
$
\begin{aligned}
& B=\{(1,2)(1,3)(1,5)(2,1)(3,1)(5,1)\} \\
& n(B)=6
\end{aligned}
$
$
P ( B )=\frac{ n ( B )}{ n (S)}=\frac{6}{36}=\frac{1}{6}
$
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Question 181 Mark
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she gets double entry fee
Answer
Sample space $( S )=\{ HHH , HHT , HTH , HTT , THH , THT , TTH , TTT \}$
$
n(S)=8
$

Let $A$ be the event of getting double entry fee ...(only getting 3 heads)
$
n(A)=1
$
$
P(A)=\frac{n(A)}{n(S)}
$
$
=\frac{1}{8}
$
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Question 191 Mark
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she just gets her entry fee
Answer
$\begin{aligned} & \text { Sample space }( S )=\{ HHH , HHT , HTH , HTT , THH , THT , TTH , TTT \} \\ & n ( S )=8 \\ & \text { Let B be the event of getting her entry fee ...(one or two heads to show) } \\ & n ( B )=\text { Probability of one head + Probability of } 2 \text { head } \\ & =\frac{3}{8}+\frac{3}{8} \\ & =\frac{6}{8} \\ & =\frac{3}{4}\end{aligned}$
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Question 201 Mark
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she loses the entry fee
Answer
Sample space $( S )=\{ HHH , HHT , HTH , HTT , THH , THT , TTH , TTT \}$
$
n(S)=8
$

To loss the entry means not getting the head ...(only tail)
$
n(C)=1
$
$
P(C)=\frac{n(C)}{n(S)}
$
$
=\frac{1}{8}
$
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Question 211 Mark
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on the same day?
Answer
$\begin{aligned} & \text { Sample space }(S)=6 \times 6=36 \\ & n(S)=36 \\ & \text { [priya and Amuthan are visiting a particular shop in any one of } 6 \text { days is } 6 \times 6=36 \text { ] } \\ & \text { Let } A \text { be the event of getting both are shopping on the same day } \\ & A=\{(\text { Mon, Mon) (Tue, Tue) (Wed, Wed) (Thu, Thu) (Fri, Fri) (Sat, Sat) }\} \\ & n ( A )=6 \\ & P(A)=\frac{n(A)}{n(S)} \\ & =\frac{6}{36} \\ & =\frac{1}{6}\end{aligned}$
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Question 221 Mark
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on different days?
Answer
Sample space $(S)=6 \times 6=36$
$
n(S)=36
$
[priya and Amuthan are visiting a particular shop in any one of 6 days is $6 \times 6=36$ ]

Let $B$ be the event of shopping in different days.
$
n(B)=36-6=30
$
$
P(B)=\frac{n(B)}{n(S)}
$
$
=\frac{30}{36}
$
$
=\frac{5}{6}
$
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Question 231 Mark
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on consecutive days?
Answer
Sample space $(S)=6 \times 6=36$
$
n(S)=36
$
[priya and Amuthan are visiting a particular shop in any one of 6 days is $6 \times 6=36$ ]
Let $C$ be the event of shopping consecutive days
$
\begin{aligned}
& C =\{(\text { Mon, Tue })(\text { Tue, Wed })(\text { Wed, Thu }) \text { (Thu, Fri) (Fri, Sat })\} \\
& n ( C )=5 \\
& P ( C )=\frac{ n ( C )}{ n ( S )} \\
& =\frac{5}{36}
\end{aligned}
$
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Question 241 Mark
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is a clavor
Answer
$\begin{aligned} & \text { King diamond }+ \text { Queen diamonds }=1+1=2 \ldots(1) \\ & \text { Queen hearts }+ \text { Jack of hearts }=1+1=2 \ldots(2) \\ & \text { Jack spade }+ \text { King of spades }=1+1=2 \ldots(3) \\ & \text { Remaining number of cards }=52-(6) \\ & n(S)=46 \\ & \text { Let } A \text { be the event of getting a clavor } \\ & n(A)=13 \\ & P(A)=\frac{n(A)}{n(S)}=\frac{13}{36}\end{aligned}$
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Question 251 Mark
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is a queen of red card
Answer
King diamond + Queen diamonds $=1+1=2 \quad \ldots(1)$
Queen hearts + Jack of hearts $=1+1=2 \ldots(2)$
Jack spade + King of spades $=1+1=2 \ldots(3)$
Remaining number of cards $=52-(6)$
$n(S)=46$

Let $B$ be the event of getting a queen of red card
$n(B)=2$

But the above two cards are removed from (1) and (2)
$n(B)=0$
$P(B)=\frac{n(B)}{n(S)}=\frac{0}{46}=0$
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Question 261 Mark
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is a king of black card
Answer
$\begin{aligned} & \text { King diamond + Queen diamonds }=1+1=2 \ldots(1) \\ & \text { Queen hearts + Jack of hearts }=1+1=2 \ldots(2) \\ & \text { Jack spade + King of spades }=1+1=2 \ldots(3) \\ & \text { Remaining number of cards }=52-(6) \\ & n(S)=46 \\ & \text { Let } B \text { be the event of getting a king of black card } \\ & n(B)=(2-1) \ldots[\text { from (3) one black card is removed] } \\ & n(B)=1 \\ & P(B)=\frac{n(B)}{n(S)}=\frac{1}{46}\end{aligned}$
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Question 271 Mark
The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer
Standard deviation of a data $(\sigma)=6.5$

Mean of the data $(\bar{x})=12.5$
Coefficient of variation $=\frac{\sigma}{\bar{x}} \times 100 \%$
$
\begin{aligned}
& =\frac{6.5}{12.5} \times 100 \% \\
& =52 \%
\end{aligned}
$
Coefficient of variation $=52 \%$
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Question 281 Mark
If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation
Answer
If the standard deviation of a data is 4.5 and each value of the data decreased by 5, the new standard deviation does not change and it is also 4.5.
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Question 291 Mark
Calculate the range of the following data.
Income 400 − 450 450 − 500 500 − 550 550 − 600 600 − 650
Number of workers 8 12 30 21 6
Answer
Smallest value (S) = 400
Largest value (L) = 650
Range = L – S
= 650 – 400
= 250
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Question 301 Mark
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Answer
If the range = 36.8 and
The smallest value = 13.4 then
The largest value = L = R + S
= 36.8 + 13.4
= 50.2
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Question 311 Mark
Find the range and coefficient of range of the following data
63, 89, 98, 125, 79, 108, 117, 68
Answer
$\begin{aligned} & \text { Here the largest value }(L)=125 \\ & \text { The smallest value }(S)=63 \\ & \text { Range }=L-S=125-63=62 \\ & \text { Coefficient of range }=\frac{L-S}{L+S} \\ & =\frac{125-63}{125+63} \\ & =\frac{62}{188} \\ & =0.33\end{aligned}$
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Question 321 Mark
Find the range and coefficient of range of the following data
43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Answer
The largest value $( L )=61.4$
The smallest value $(S)=13.6$
$
\begin{aligned}
& \text { Range }= L - S \\
& =61.4-13.6 \\
& =47.8 \\
& \text { Coefficient of range }=\frac{ L - S }{ L + S } \\
& =\frac{61.4-13.6}{61.4+13.6} \\
& =\frac{47.8}{75} \\
& =0.64
\end{aligned}
$
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[1 Mark Questions] - MATHS STD 10 Questions - Vidyadip