CBSE Class 10 Mathematics Probability VBQs Set C

Short Answer Type Questions-

Question. A game consists of tossing a one-rupee coin 3 times and noting the outcome each time. Ramesh wins the game if all the tosses give the same result (i.e., three heads or three tails) and loses otherwise. Find the probability of Ramesh losing the game.
Answer: On tossing a coin three times,
Possible outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒ Total number of outcomes = 8
Favourable outcomes = {HHT, HTH, THH, HTT, THT, TTH}
⇒ Number of favourable outcomes = 6
∴ Probability of losing the game \( = \frac{\text{No. of favourable outcomes}}{\text{Total no. of outcomes}} = \frac{6}{8} = \frac{3}{4} \).

Question. A bag contains 14 balls of which x are white. If 6 more white balls are added to the bag, the probability of drawing a white ball is \( \frac{1}{2} \). Find the value of x.
Answer: Given : Total number of balls = 14
Number of white balls = x
When 6 more white balls are added to the bag, then
Total balls = 14 + 6 = 20
Number of white balls = x + 6
So, Probability of drawing white ball \( = \frac{\text{Number of white balls}}{\text{Total no. of balls}} = \frac{x + 6}{20} \)
But required probability \( = \frac{1}{2} \) [Given]
∴ \( \frac{x + 6}{20} = \frac{1}{2} \ r Arr 2(x + 6) = 20 \ r Arr x + 6 = 10 \ r Arr x = 4 \).

Question. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.
Answer: Let the number of blue balls be x.
Thus, total number of possible outcomes = 5 + x
Now, number of favourable outcomes if the red ball is drawn = 5.
∴ \( P(R) = \frac{5}{5 + x} \)
Again, number of favourable outcomes if the blue ball is drawn = x.
∴ \( P(B) = \frac{x}{5 + x} \)
Now, \( P(B) = 3P(R) \) [Given]
⇒ \( \frac{x}{5 + x} = 3 \left( \frac{5}{5 + x} \right) \)
⇒ x = 15
Hence, the number of blue balls = 15.

Question. A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be :
(i) a blue card,
(ii) not a yellow card, and
(iii) neither yellow nor a blue card.
Answer: Total number of cards = 100 + 200 + 50 = 350
(i) Possible no. of outcomes of drawing a blue card = 50
Hence, probability \( = \frac{50}{350} = \frac{1}{7} \).
(ii) Possible no. of outcomes of not drawing a yellow card = 350 – 200 = 150.
Hence, probability \( = \frac{150}{350} = \frac{3}{7} \).
(iii) Possible no. of outcomes of drawing neither yellow nor a blue card = 350 – (200 + 50) = 350 – 250 = 100
Hence, probability \( = \frac{100}{350} = \frac{2}{7} \).

Question. A lot consists of 144 ball pens of which 20 are defective. The customer will buy a ball pen, if it is good, but will not buy a defective ball pen. The shopkeeper draws one pen at random from the lot and gives it to the customer. What is the probability that
(i) customer will buy the pen
(ii) customer will not buy the pen
Answer: Total number of pens = 144
No. of defective pens = 20
⇒ No. of non-defective pens = 144 – 20 = 124
(i) A customer will buy the pen, if it is non-defective.
So, probability of drawing non-defective pen \( = \frac{\text{No. of non-defective pens}}{\text{Total no. of pens}} = \frac{124}{144} = \frac{31}{36} \).
(ii) A customer will not buy the pen, if it is defective.
So, probability of drawing defective pen \( = \frac{\text{No. of defective pens}}{\text{Total no. of pens}} = \frac{20}{144} = \frac{5}{36} \).

Question. Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one dice and squares the number that appears on it. Who has the better chance of getting the number 36 ? Why ?
Answer: Apoorv throws two dice once.
So, total number of outcomes, \( n(S_1) = 36 \).
Number of outcomes for getting product 36,
\( n(E_1) = \{(6, 6)\} = 1 \)
∴ \( P_1 = \frac{n(E_1)}{n(S_1)} = \frac{1}{36} \).
Also, Peehu throw one dice.
So, total number of outcomes, \( n(S_2) = 6 \).
Number of outcomes for getting square of a number is 36,
\( n(E_2) = 1 \) [\( \because 6^2 = 36 \)]
∴ \( P_2 = \frac{n(E_2)}{n(S_2)} = \frac{1}{6} = \frac{6}{36} \).
\( \because P_2 > P_1 \)
Hence, Peehu has better chance of getting the number 36.

Question. An unbiased die is rolled twice. Find the probability of getting (i) the sum of two numbers as a prime and (ii) the sum of two numbers equal to 9.
Answer: (i) The sum of the two numbers lies between 2 and 12. So the required primes are 2, 3, 5, 7, 11.
No. of ways for getting 2 = {(1, 1)} = 1
No. of ways for getting 3 = {(1, 2)}, {(2, 1)} = 2
No. of ways of getting 5 = {(1, 4) , (4, 1), (2, 3), (3, 2)} = 4
No. of ways of getting 7 = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} = 6
No. of ways of getting 11= {(5, 6), (6, 5)} = 2
No. of favourable ways = 1 + 2 + 4 + 6 + 2 = 15
Total ways = \( 6 \times 6 = 36 \)
Probability of getting a sum as a prime = \( \frac{15}{36} = \frac{5}{12} \)
(ii) No. of ways of getting a sum of 9 = {(3, 6), (6, 3), (4, 5), (5, 4)} = 4
Probability of getting a sum of 9 = \( \frac{4}{36} = \frac{1}{9} \)

Question. A jar contains only green, white and yellow marbles. The probability of selecting a green marble and white marble randomly from a jar is 1/4 and 1/3 respectively. If this jar contains 10 yellow marbles, what is the total number of marbles in the jar ?
Answer: Let the no. of green marbles = \( x \)
Let the no. of white marbles = \( y \)
∴ Total no. of marbles = \( x + y + 10 \)
Now, \( P(\text{selecting a green marble}) = \frac{x}{x + y + 10} \)
But it is given that, \( P(\text{green marble}) = \frac{1}{4} \)
∴ \( \frac{x}{x + y + 10} = \frac{1}{4} \)
\( \Rightarrow 4x = x + y + 10 \)
\( \Rightarrow 3x - y = 10 \text{ ...(i)} \)
Similarly,
\( P(\text{selecting a white marble}) = \frac{y}{x + y + 10} \)
∴ \( \frac{y}{x + y + 10} = \frac{1}{3} \)
\( 3y = x + y + 10 \)
\( \Rightarrow 2y - x = 10 \text{ ...(ii)} \)
On solving equations (i) and (ii), we get
\( x = 6 \) and \( y = 8 \).
∴ Total no. of marbles = \( 6 + 8 + 10 = 24 \).

Question. Find the probability of getting 53 Fridays in a leap year.
Answer: Total number of days in a leap year = 366
Total number of days in a week = 7
Thus, the number of weeks in a leap year = \( \frac{366}{7} = 52 \text{ weeks 2 days} \)
Now, the two days can be Sunday and Monday; Monday and Tuesday; Tuesday and Wednesday; Wednesday and Thursday; Thursday and Friday; Friday and Saturday; Saturday and Sunday.
So, total no. of possible outcomes = 7
And the favourable outcomes are Thursday and Friday; Friday and Saturday
So, no. of sum of the favourable outcomes = 2
∴ Probability = \( \frac{2}{7} \).

Question. From a pack of 52 playing cards, Jacks, Queens and kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is : (i) a black king (ii) a card of red colour (iii) a card of black colour
Answer: Since, Jacks, Queens and Kings of red colour are removed. Then,
Total number of possible outcomes = 52 – 6 = 46
(i) Let \( E_1 \) be the event of getting a black king.
∴ Favourable outcomes = king of spade and king of club.
\( \Rightarrow \) No. of favourable outcomes = 2
\( P(E_1) = \frac{2}{46} = \frac{1}{23} \)
(ii) Let \( E_2 \) be the event of getting a card of red colour
∴ Favourable outcomes = 10 cards of heart and 10 cards of diamond.
No. of favourable outcomes = 20
\( P(E_2) = \frac{20}{46} = \frac{10}{23} \)
(iii) Let \( E_3 \) be the event of getting a card of black colour
∴ Favourable outcomes = 13 cards of spade and 13 cards of club.
\( \Rightarrow \) No. of favourable outcomes = 26
\( P(E_3) = \frac{26}{46} = \frac{13}{23} \)

Question. From a pack of 52 cards, Aces, Queens and kings are removed. The remaining cards are well-shuffled and then a card is drawn from it. Find the probability that the drawn card is (i) a black face card, and (ii) a red card.
Answer: Total number of cards = 52
Number of cards removed = 4 + 4 + 4 = 12
Hence, total number of remaining cards = 52 – 12 = 40.
(i) Number of black face cards (Jack card) = 1 + 1 = 2
Hence, Probability = \( \frac{2}{40} = \frac{1}{20} \)
(ii) Number of red cards = 13 hearts + 13 of diamond – (1 king of hearts + 1 king of diamonds + 1 queen of hearts + 1 queen of diamonds + 1 ace of hearts + 1 ace of diamonds) = 20
Hence, Probability = \( \frac{20}{40} = \frac{1}{2} \).

Question. Cards marked with numbers 1, 2, 3, 4, …, 20 are well-shuffled and a card is drawn from it at random. What is the probability that the number on the card is (i) a prime number, (ii) divisible by 3, and (iii) a perfect square.
Answer: Total no. of outcomes = 20.
(i) The favourable outcomes for a prime number are 2, 3, 5, 7, 11, 13, 17, 19.
Thus, no. of favourable outcomes = 8.
∴ Probability = \( \frac{8}{20} = \frac{2}{5} \)
(ii) The most favourable outcomes for a number divisible by 3 are 3, 6, 9, 12, 15, 18.
Thus, no. of most favourable outcomes for a number divisible by 3 = 6.
∴ \( P(E) = \frac{6}{20} = \frac{3}{10} \)
(iii) The most favourable outcomes for a perfect square are 1, 4, 9, 16.
Thus, no. of most favourable outcomes for a perfect square = 4.
∴ \( P(E) = \frac{4}{20} = \frac{1}{5} \).

Passage Based Questions

Question. Read the following passage and answer the questions that follows. A teacher wrote some numbers on black board. These number are. (a) 0.94 (b) 0.97 (c) – 0.3 (d) 0.5 (e) 1.5 (f) 0.99

Question. Out of the following, which numbers cannot be the probability of an event?
Answer: – 0.3, 1.5

Question. What is the probability of a sure event?
Answer: The probability of a sure event is always 1.

Long Answer Type Questions

Question. The King, Queen and Jack of clubs are removed from a pack of 52 cards and then the remaining cards are well-shuffled. A card is selected from the remaining cards. Find the probability of getting a card (i) of spade (ii) of black king (iii) of clubs (iv) of jacks
Answer: Total number of cards = 52
Number of cards removed = 3
So, Number of remaining cards = 52 – 3 = 49
(i) Possible number of outcomes of drawing a spade = 13
∴ \( P(E) = \frac{13}{49} \)
(ii) Possible number of outcomes of drawing a black king = 1
∴ \( P(E) = \frac{1}{49} \)
(iii) Possible number of outcomes of drawing a club = 13 – 3 = 10
∴ \( P(E) = \frac{10}{49} \)
(iv) Possible number of outcomes of drawing a jack = 4 – 1 = 3
∴ \( P(E) = \frac{3}{49} \)

Question. From a pack of 52 playing cards, Jacks and Kings of red colour and Queens and Aces of black colour are removed. The remaining cards are mixed and a card is drawn at random. Find the probability that the card drawn is (i) a black Queen (ii) a card of red colour (iii) a Jack of black colour (iv) a face card
Answer: Total number of cards = 52
Number of cards removed = 4 + 4 = 8
So, Number of remaining cards = 52 – 8 = 44
(i) Number of possible outcomes of drawing a black queen = 0
[\(\because\) queens of black colour are removed]
∴ \( P(E) = 0 \)
(ii) Number of possible outcomes of drawing a red coloured card = 26 – 4 = 22
∴ \( P(E) = \frac{22}{44} = \frac{1}{2} \)
(iii) Number of possible outcomes of drawing a jack of black colour = 2
∴ \( P(E) = \frac{2}{44} = \frac{1}{22} \)
(iv) Number of possible outcomes of drawing a face card = 2 black jacks + 2 black kings + 2 red queens = 6
∴ \( P(E) = \frac{6}{44} = \frac{3}{22} \)

Question. Five cards the ten, jack, queen, king and ace of diamonds are well shuffled with their face downwards. One card is then picked up in random. (i) What is the probability that the drawn card is a queen? (ii) If the queen is drawn and put aside and a second card is drawn, find the probability that the second card is (a) an ace and (b) a queen.
Answer: Total number of cards = 5
(i) Possible outcomes of drawing a queen = 1
Hence, \( P(E) = \frac{1}{5} \)
(ii) Now the queen is drawn and set aside.
So, the total number of cards = 4
(a) Possible outcomes of drawing an ace = 1
Hence, \( P(E) = \frac{1}{4} \)
(b) Possible outcomes of drawing a queen = 0
Hence, \( P(E) = 0 \)
This is an impossible event.

Question. Find the probability that a number selected at random from the numbers 1, 2, 3, 4, …, 34, 35 is a (i) prime, (ii) multiple of 7 and (iii) divisible by 3 or 5.
Answer: All possible outcomes = {1, 2, 3, …, 34, 35}
∴ Total number of all possible outcomes = 35
(i) The favourable outcomes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
Thus, total number of favourable outcomes = 11
Hence, \( P(E) = \frac{11}{35} \)
(ii) The favourable outcomes = {7, 14, 21, 28, 35}
Thus total number of favourable outcomes = 5
Hence, \( P(E) = \frac{5}{35} = \frac{1}{7} \)
(iii) The favourable outcomes = {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35}
Thus, total number of favourable outcomes = 16
Hence, \( P(E) = \frac{16}{35} \).

Question. A bag contains cards, numbered from 1 to 90. A card is drawn at random from the box. Find the probability that the selected card bears a (i) two-digit number (ii) perfect square number
Answer: Total number of cards = 90
(i) Number of two-digit number = 90 – 9 = 81 [10, 11, 12, 13, ...... 90]
∴ \( P(E) = \frac{81}{90} = \frac{9}{10} \)
(ii) Perfect squares between 1 and 90 = 4, 9, 16, 25, 36, 49, 64, 81
\( \Rightarrow \) Number of perfect squares = 8
∴ \( P(E) = \frac{8}{90} = \frac{4}{45} \)

Question. A box contains cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that number on the drawn card is (i) a prime number (ii) a composite number (iii) a number divisible by 3
Answer: Total number of cards = 20
(i) Prime numbers between 1 and 20 = 2, 3, 5, 7, 11, 13, 17, 19
Thus, number of prime numbers = 8.
∴ \( P(E) = \frac{8}{20} = \frac{2}{5} \)
(ii) Composite numbers between 1 and 20 = 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20
Thus, number of composite numbers = 11.
∴ \( P(E) = \frac{11}{20} \)
(iii) Numbers divisible by 3 between 1 and 20 = 3, 6, 9, 12, 15, 18
Thus, number of outcomes divisible by 3 = 6.
∴ \( P(E) = \frac{6}{20} = \frac{3}{10} \)

Question. A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) white, (ii) red, (iii) black and (iv) not red ?
Answer: Sol. The sum of the total possible outcomes \( = 3 + 5 + 4 = 12 \)
(i) Total number of favourable outcomes for drawing a white ball = 4
∴ \( P(E) = \frac{4}{12} = \frac{1}{3} \)
(ii) Total number of favourable outcomes for drawing a red ball = 3
∴ \( P(E) = \frac{3}{12} = \frac{1}{4} \)
(iii) Total number of favourable outcomes for drawing a black ball = 5
∴ \( P(E) = \frac{5}{12} \)
(iv) Total number of favourable outcomes for drawing a ball not red \( = 5 + 4 = 9 \)
∴ \( P(E) = \frac{9}{12} = \frac{3}{4} \).

Question. A box contains 20 balls bearing numbers from 1 to 20. A ball is drawn at random from the box. Find the probability that the number on the ball is (i) an odd number, (ii) divisible by 2 or 3, (iii) not divisible by 10 and (iv) a prime number.★★
Answer: Sol. The total number of possible outcomes = 20
(i) The favourable outcomes for numbers that are odd are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Hence, the total number of favourable outcomes = 10
∴ \( P(E) = \frac{10}{20} = \frac{1}{2} \)
(ii) The favourable outcomes for numbers divisible by 2 or 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20.
Hence, the total number of favourable outcomes = 13
∴ \( P(E) = \frac{13}{20} \)
(iii) The favourable outcomes for numbers that are not divisible by 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19.
Hence, the total number of favourable outcomes = 18.
∴ \( P(E) = \frac{18}{20} = \frac{9}{10} \)
(iv) The favourable outcomes for prime numbers are 2, 3, 5, 7, 11, 13, 17, 19.
Hence, the total number of favourable outcomes = 8
∴ \( P(E) = \frac{8}{20} = \frac{2}{5} \).

Question. A box contains 19 balls bearing numbers from 1 to 19. A ball is drawn at random from the box. Find the probability that the number on the ball is (i) a prime number, (ii) divisible by 3 or 5, (iii) neither divisible by 5 nor 10, and (iv) an even number.
Answer: Sol. The total number of possible outcomes = 19
(i) The favourable outcomes for prime numbers are 2, 3, 5, 7, 11, 13, 17, 19.
Hence, total number of favourable outcomes = 8
∴ \( P(E) = \frac{8}{19} \)
(ii) The favourable outcomes for numbers divisible by 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18.
Hence, total number of favourable outcomes = 8
∴ \( P(E) = \frac{8}{19} \)
(iii) The favourable outcomes for numbers that are neither divisible by 5 nor 10 are 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19.
Hence, total number of favourable outcomes = 16
∴ \( P(E) = \frac{16}{19} \)
(iv) The favourable outcomes for numbers that are even are 2, 4, 6, 8, 10, 12, 14, 16, 18.
Hence, total number of favourable outcomes = 9
∴ \( P(E) = \frac{9}{19} \).

Question. Two different dice are thrown together. Find the probability that the numbers obtained (i) have a sum less than 7. (ii) have a product less than 16. (iii) is a doublet of odd numbers.
Answer: Sol. Total number of possible outcomes in each case \( = 6 \times 6 = 36 \)
(i) Having a sum less than 7
Possible outcomes are, (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 1), (5, 1)
∴ \( n(E) = 15 \)
So, probability \( = \frac{25}{36} \)
(ii) Having a product less than 16
Possible outcomes are, (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), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2),
∴ \( n(E) = 25 \)
So, probability \( = \frac{25}{36} \)
(iii) A doublet of odd numbers
Possible outcomes are (1, 1), (3, 3), (5, 5)
∴ \( n(E) = 3 \)
So, probability \( = \frac{3}{36} = \frac{1}{12} \).

Question. A bag contains 7 red balls, 5 white balls, 2 blue balls and 4 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn out from the bag is (i) white or blue, (ii) red or black, (iii) not white and (iv) neither white nor black.
Answer: Sol. Total number of outcomes \( = 7 + 5 + 2 + 4 = 18 \)
(i) Number of favourable outcomes for white or blue balls \( = 5 + 2 = 7 \)
∴ \( P(E) = \frac{7}{18} \)
(ii) Number of favourable outcomes for red or black balls \( = 7 + 4 = 11 \)
∴ \( P(E) = \frac{11}{18} \)
(iii) Number of favourable outcomes for balls not white \( = 7 + 4 + 2 = 13 \)
∴ \( P(E) = \frac{13}{18} \)
(iv) Number of favourable outcomes for balls neither white nor black \( = 7 + 2 = 9 \)
∴ \( P(E) = \frac{9}{18} = \frac{1}{2} \).

Question. A bag contains 4 red, 6 white and 5 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn out from the bag is (i) white, (ii) red, (iii) not black and (iv) red or white.
Answer: Sol. Total number of outcomes \( = 4 + 6 + 5 = 15 \)
(i) Number of favourable outcomes for white balls = 6
∴ \( P(E) = \frac{6}{15} = \frac{2}{5} \)
(ii) Number of favourable outcomes for red balls = 4
∴ \( P(E) = \frac{4}{15} \)
(iii) Number of favourable outcomes for balls not black \( = 4 + 6 = 10 \)
∴ \( P(E) = \frac{10}{15} = \frac{2}{3} \)
(iv) Number of favourable outcomes for red or white balls \( = 4 + 6 = 10 \)
∴ \( P(E) = \frac{10}{15} = \frac{2}{3} \).

Question. A die is rolled twice. Find the probability that (i) 5 will not come up either time and (ii) 5 will come up exactly once.
Answer: Sol. Total possible outcomes = (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)
Thus, total number of possible outcomes = 36
(i) Now, the total number of favourable outcomes = 25
Hence, \( P(E) = \frac{25}{36} \)
(ii) The favourable outcomes are (1, 5); (2, 5); (3, 5); (4, 5); (6, 5); (5, 1); (5, 2); (5, 3); (5, 4); (5, 6)
Thus, the total number of favourable outcomes = 10
Hence, \( P(E) = \frac{10}{36} = \frac{5}{18} \).

Assertion and Reasoning Based Questions

DIRECTIONS : In the following questions, a statement 1 is followed by statement 2. Mark the correct choice as :
(A) If both statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1.
(B) If both statement 1 and statement 2 are true but statement 2 is not the correct explanation of statement 1.
(C) If statement 1 is true, but statement 2 is false.
(D) If statement 1 is false and statement 2 is true.

Question. Statement 1 : In rolling a dice, the probability of getting number 9 is zero.
Statement 2 : It is an impossible event.
Answer: (A) From the statement 1, the probability of getting 9 in rolling a die is an impossible event as the sample space for rolling a die is 6. The probability of getting 9 in rolling a die is 0. Therefore, the given statement 1 and statement 2 and true and statement 2 is the correct explanation of statement 1.

Question. Statement 1 : If the probability is 0.0001 then the event is very unlikely to happen.
Statement 2 : If \( P(A) \) denotes the probability of an event \( A \), then \( 0 \le P(A) \le 1 \).
Answer: (B) Statement 1 is about the unlike event and in statement 2 the condition for probability is given. So both the statements are true individually. Both Statement 1 and statement 2 are true but statement 2 is not correct explanation of statement 1.

Question. Statement 1 : The probability of the complementary event is \( 1 - P \), if the probability of an event is \( P \).
Statement 2 : \( P(E) + P(\bar{E}) = 1 \) If \( E \) and \( \bar{E} \) are complementary events.
Answer: (A) In the statement 1, it is given about the complementary event which is true and in statement 2, another condition for complementary events is given which is also true. \( \therefore \) It is the correct explanation of statement 1.