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 $\frac{1}{200}$

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