ML AGGARWAL CLASS 8 Chapter 19 Data Handiling Exercise 19.3

  Exercise 19.3

Question 1

Sol :

(i) A, A, A, B, C, D 

(ii)W , R, B, G, Y 


Question 2

Sol :

(i) Total No. of Outcomes = $\{1,2,3,4,5,6\}=6$

Favorable Outcomes = $\{2,4,6\}=3$

Probability $=\frac{\text { no.of  favorable outcomes }}{\text { Total no.of outcomes }}$

$\begin{aligned} & \frac{3}{6} \\=& 1 / 2 \end{aligned}$


(ii)  Total No. of Outcomes = $\{1,2,3,4,5,6\}=6$

Favorable Outcomes= $\{3,6\}=2$

Probability $=\frac{\text { no.of  favorable outcomes }}{\text { Total no.of outcomes }}$

=$\frac{2}{6}$

$=\frac{1}{3}$


(iii) No. of multiple of '3' = {1,2,4,5}= 4

Probability = $\frac{4}{6}=\frac{2}{3}$



Question 3

Sol :

(i) Total Outcomes = $\{T T, TH, H T, H H\}=4$

Getting two tail = {T,T } = 1

Probability $=\frac{1}{4}$


(ii) At least One tail = {TH, HT, TT}= 3

Probability $=\frac{\text { no.of  favorable outcomes }}{\text { Total no.of outcomes }}$

$=\frac{3}{4}$


(iii) No. tail = {H,H }= 1

Probability = $\frac{1}{4}$



Question 4

Sol :

Total outcomes = {TTT, TTH , THT , HTT, THH, HHT,HTH, HHH}= 8

(i) At least two heads 

Favorable outcomes = {TTT,TTH , THT, HTT} = 4

Probability $=\frac{4}{8}=\frac{1}{2}$


(ii) At least on tail 

Favorable Outcomes = {TTT, TTH, THT, HTTT, THH, HHT,HTH}= 7

Probability = $\frac{7}{8}$


(iii) At most one tail

Favorable Outcomes = {HHH,THH, HTH,HHT}= 4

Probability $=\frac{4}{8}=\frac{1}{2}$


Question 5

Sol :

When two dice rolled simultanilouly 

Total no. of outcomes =36

(i) The sum as 7 

Favorable Outcomes = {(11,6),(2,5),(3,4),(4,3),(5,2),(6,11)}

Probability $=\frac{6}{36}=\frac{1}{6}$


(ii) The sum as 3 or 4

Sum '3' = {(1,2),(2,1)}= 2

Sum '4' = {(1,3),(2,2),(3,1)}=3

Total = 2+3= 5

∴ Favorable Outcomes = 5

Probability = $\frac{5}{36}$



(iii) Prime number on both dice 

Favorable Outcomes = {(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5)}=9

probability $=\frac{9}{36}=\frac{1}{4}$



Question 6

Sol :

Total Screws = 60 

Rusted Screws= $600 \times \frac{1}{10}=60$


(i) A rusted Screw 

No. of favorable outcomes = 60

Probability=$\frac{60}{600}=\frac{1}{10}=0.1$

(ii) Not a Rusted Screw 

No of favorable outcomes = 600-60 = 540

Probability = $\frac{540}{60}=0.9$



Question 7

Sol :

TRIANGLE 

Vowels ----{I, A, E }= 3

Total letters = 8

Probability = $\frac{\text { no.of vowels }}{\text { total no. of letter}}$

$=\frac{3}{8}$



Question 8

Sol :

Bag = {5 Red, 6 black, 4 white } = 15 Balls 

(i) Getting white 

No. of favorable outcomes = 4

probability  $=\frac{4}{15}$


(ii) Not black 

Mean either red or white 

No. of favorable outcomes = 5+ 4= 9

probability=$\frac{9}{15}=\frac{3}{5}$


(iii) Red or black 

No. of favorable Outcomes = 5+ 6= 11

Probability $4=\frac{11}{15}$



Question 9

Sol :

Total Cards = 17 = {1,2,3,4,5,6.............10,11,12,13,14,15,16,17}

(i) odd

No. of favorable Outcomes = 9

Probability $=\frac{9}{17}$


(ii) Even 

No. of favorable outcomes = 8 

Probability $\frac{8}{17}$


(iii) Prime 

No. f favorable Outcomes {2,3,5,9,11,13,17}= 7

probability $=\frac{7}{17}$


(iv) divisible by 3

No of favorable outcomes = {3,6,9,12,15}=5

Probability $=\frac{5}{17}$


(v) Divisible by 2 and 3 both

No.of Favorable Outcomes = {6,12}=2

Probability $=\frac{2}{17}$


Question 10

Sol :

Total Cards = 52


(i) An ace 

No. of favorable Outcomes = 4

$\begin{aligned} \text { probability }=& \frac{4}{52} \\ &=\frac{1}{13} \end{aligned}$


(ii) A red card

No.of favorable Outcomes = 26

$\begin{aligned} \text { Probability } &=\frac{26}{52} \\ &=\frac{1}{2} \end{aligned}$


(iii) Neither a king nor a queen 

No. of favorable outcome = $56-(4 \times 2)=56-8 =44$

Probability = $\frac{44}{52}$

=$\frac{11}{13}$


(iv) A Red face card 

 No.of favorable outcome= $3 \times 2=6$\

$\begin{aligned} \text { Probability } &=\frac{6}{52} \\ &=\frac{3}{26} \end{aligned}$


(v) A card of spade or an ace 

No. of spades $=n(s)=13$

No.of Ace = n(A) = 4

No.of Ace or spade = $n(s \wedge A)=1$

$\begin{aligned} \therefore n(S \cup A) &=n(s)+n(A)-n (S \wedge A) \\ &=13+4-1 \end{aligned}$

$n(S \cup A)=16$

No.of favorable outcomes = 16

$\begin{aligned} \text { Probabilily } &=\frac{16}{54} \\ &=\frac{4}{13} \end{aligned}$


(vi) Non face card of red colour

No.of favorable outcomes = $26-(3 \times 2)=20$

Probability $=\frac{20}{54}$

$\frac{5}{13}$



Question 11

Sol :

Total tickets =5+955=1000

No.of favorable outcomes = 5

Probability = $\frac{5}{1000}$

$=\frac{1}{200}$

∴ Probability that the person wins lottery is $\frac{1}{200}$

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more