Monday always follows Sunday in the calendar, so (i) is certain. Mumbai in July is hot and monsoon-season, not a place where snowfall normally occurs, so (ii) is impossible in ordinary school-level reasoning. An elephant entering a classroom is possible only in an unusual situation, so (iii) is less likely. Greeting a friend at school tomorrow is very likely if school is open and friends are present, but not logically guaranteed.
(i) Certain; probability $1$.
(ii) Impossible or almost impossible in ordinary experience; probability close to $0$.
(iii) Less likely, practically close to $0$.
(iv) More likely; probability close to $1$, though not certain.
In the sample, 8 out of 30 sweets are green, so the experimental probability of green is $\dfrac{8}{30}=\dfrac{4}{15}$. Yellow sweets form $\dfrac{7}{30}$ of the sample, so in 600 sweets the estimate is $\dfrac{7}{30}\times600=140$.
(i) $\dfrac{8}{30}=\dfrac{4}{15}$
(ii) About $140$ yellow sweets.
Arts Club was preferred by 11 out of 40 students, so the probability is $\dfrac{11}{40}$. Sports Club was preferred by 9 out of 40 students, so the estimate for 800 students is $\dfrac{9}{40}\times800=180$.
(i) $\dfrac{11}{40}$
(ii) About $180$ students.
There are $6$ possible outcomes: $\{1,2,3,4,5,6\}$.
For a combined die-and-coin experiment, pair each die outcome with each coin outcome. Between $-5$ and $+5$ inclusive, list every integer. For drawing one ball when only colour is recorded, the possible colour outcomes are green and red.
(i) $S=\{(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T),(5,H),(5,T),(6,H),(6,T)\}$.
(ii) $S=\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$.
(iii) By colour, $S=\{\text{green},\text{red}\}$.
Each snack can be paired with either of the 2 drinks, so there are $3\times2=6$ combinations. The Samosa event contains the two outcomes where the snack is Samosa.
(i) $S=\{(\text{Samosa},\text{Chai}),(\text{Samosa},\text{Lassi}),(\text{Pakora},\text{Chai}),(\text{Pakora},\text{Lassi}),(\text{Bhaji},\text{Chai}),(\text{Bhaji},\text{Lassi})\}$.
(ii) Event $=\{(\text{Samosa},\text{Chai}),(\text{Samosa},\text{Lassi})\}$.
Basket A has 3 fruits, of which 1 is an apple, so $P(\text{Apple})=1/3$. Basket B has 2 fruits, of which 1 is a banana, so $P(\text{Banana})=1/2$. The choices are independent, so the probability of apple and banana is $1/3\times1/2=1/6$.
(i) The tree branches from Basket A to Apple or Orange, and from each of these to Banana or Mango.
(ii) By fruit type, $S=\{(\text{Apple},\text{Banana}),(\text{Apple},\text{Mango}),(\text{Orange},\text{Banana}),(\text{Orange},\text{Mango})\}$.
(iii) $P(\text{Apple and Banana})=\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{1}{6}$.
Because the pen is replaced, both picks have the same colour probabilities: red $3/9$, black $4/9$, green $2/9$. Same colour can happen as RR, BB or GG. Add their probabilities: $\dfrac{3}{9}\cdot\dfrac{3}{9}+\dfrac{4}{9}\cdot\dfrac{4}{9}+\dfrac{2}{9}\cdot\dfrac{2}{9}=\dfrac{9+16+4}{81}=\dfrac{29}{81}$.
(i) Colour-pair outcomes are $\{RR,RB,RG,BR,BB,BG,GR,GB,GG\}$, where R = red, B = black and G = green.
(ii) The probability of the same colour is $\left(\dfrac{3}{9}\right)^2+\left(\dfrac{4}{9}\right)^2+\left(\dfrac{2}{9}\right)^2=\dfrac{29}{81}$.
(i) $0$
(ii) sample space
(iii) $1$
(iv) $\dfrac{1}{2}$
Frequency is the count of times an outcome occurs. Relative frequency is the count divided by the total number of observations.
The number 15 is the frequency, and the relative frequency is $\dfrac{15}{50}=\dfrac{3}{10}=0.3$.
Equally likely outcomes require symmetry or equal counts with no bias. A fair coin has H and T equally likely, and a fair die has all six faces equally likely. A car starting depends on its condition. A bag with 3 red and 7 blue marbles gives unequal colour probabilities. Birth outcomes may be close, but they are not assumed exactly equal in the same theoretical way as a fair coin.
(i) Not equally likely.
(ii) Equally likely.
(iii) Equally likely.
(iv) Not equally likely if outcomes are recorded by colour; blue is more likely than red.
(v) Not treated as exactly equally likely without data; it is not a designed fair random device like a coin or die.
Probability is favourable outcomes divided by total equally likely outcomes. For two coins, at least one head means HH, HT or TH. For cards 1 to 10, even cards are 2,4,6,8,10. For one die, numbers greater than 4 are 5 and 6. For the balls, not red means blue or green, $2+1=3$ balls out of 6. For three coins, exactly two heads are HHT, HTH and THH.
(i) $S=\{HH,HT,TH,TT\}$; probability $=\dfrac{3}{4}$.
(ii) $S=\{1,2,3,4,5,6,7,8,9,10\}$; probability $=\dfrac{5}{10}=\dfrac12$.
(iii) $S=\{1,2,3,4,5,6\}$; probability $=\dfrac{2}{6}=\dfrac13$.
(iv) By individual balls, total outcomes $=6$; not red outcomes $=3$; probability $=\dfrac12$.
(v) $S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$; probability $=\dfrac{3}{8}$.
There are 3 equally likely candies and 1 favourable outcome, strawberry. Therefore the probability is $1/3$.
$\dfrac{1}{3}$.
Each of the 2 shirts can be paired with each of the 3 pants, so there are $2\times3=6$ combinations. A table would have rows Red and Blue, and columns Jeans, Khakis and Shorts.
The 6 outfits are: red shirt with jeans, red shirt with khakis, red shirt with shorts, blue shirt with jeans, blue shirt with khakis, and blue shirt with shorts.