CBSE Class 10 Probability Sure Shot Questions Set 02

CBSE Class 10 Probability Sure Shot Questions Set B. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

1. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(a) an ace card

(b) a red card

(c) either red or king card

(d) red and a king

(e) ‘2’ of spades

(f) ‘10’ of a black suit

(g) a queen of black suit

(h) either a black card or a king

(i) black and king card

(j) a jack, queen or a king

(k) a heart card

(l) a queen card

(m) the ace of spades

(n) the seven of clubs

(o) a ten

(p) a black card

(q) neither a heart nor a king

(r) neither an ace nor a king

(s) neither a red card nor a queen card

(t) a face card or an ace

(u) a face card or a black card

(v) a face card and a black card

(w) neither a face card nor an ace

(x) neither a face card nor ’10’ card

(y) either a king or red card

(z) either an ace or black card

(aa) an ace and a black card

(bb) a king of red colour card

(cc) a face card

(dd) a red face card

(ee) the jack of hearts

(ff) a spade card

(gg) the queen of diamonds

(hh) ‘9’ of black suit

(ii) a face card or spade card

2. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is

(a) an ace? (b) a queen? 

3. The king , queen and jack of clubs are removed from a pack of 52 playing cards. One card is selected at random from the remaining cards. Find the probability that the card is

(a) neither a heart nor a king

(b) neither an ace nor a king

(c) neither a red card nor a queen card

(d) a black card or an ace.

(e) either a heart or a spade card

(f) a king card

(g) a heart card

(h) a red card

(i) a black card

(j) a spade card

(k) a diamond card

(l) a club card

(m) either an ace card or black card

(n) an ace card

(o) a face card

(p) a face card with red colour

(q) neither ‘10’ card nor an ace

(r) an even number card

(s) an odd number card

(t) not a natural number.

6. All spades are removed from a well shuffled deck of 52 cards and then one card is drawn randomly from the remaining cards. Find the probability of getting

(a) neither a heart nor a king

(b) neither an ace nor a king

(c) neither a red card nor a queen card

(d) a black card or an ace.

(e) either a heart or a spade card

(f) a red card

(g) a black card

(h) a spade card

(i) a diamond card

(j) a club card

(k) either an ace card or black card

(l) an ace card

(m) a face card with red colour

(n) neither ‘10’ card nor an ace

(o) an even number card

(p) a face card

(q) an odd number card

4. All face cards are removed from a well shuffled deck of 52 cards and then one card is drawn randomly from the remaining cards. Find the probability of getting.

(a) neither a heart nor a king

(b) neither an ace nor a king

(c) neither a red card nor a queen card

(d) a black card or an ace.

(e) either a heart or a spade card

(f) a red card

(g) a black card

(h) a spade card

(i) a diamond card

(j) a club card

(k) either an ace card or black card

(l) an ace card

(m) a face card with red colour

(n) neither ‘10’ card nor an ace

(o) an even number card

(p) an odd number card

5. All cards of ace, jack and queen are removed from a deck of playing cards. One card is drawn at random from the remaining cards, find the probability that the card drawn

(a) neither a heart nor a king

(b) neither an ace nor a king

(c) neither a red card nor a queen card

(d) a black card or an ace.

(e) either a heart or a spade card

(f) a king card

(g) a heart card

(h) a red card

(i) a black card

(j) a spade card

(k) a diamond card

(l) a club card

(m) either an ace card or black card\

(n) an ace card

(o) a face card

(p) a face card with red colour

(q) neither ‘10’ card nor an ace

(r) an even number card

(s) an odd number card

(t) not a natural number.

6. All cards of ‘10’, an ace and queen cards are removed from a well shuffled deck of 52 cards and then one card is drawn randomly from the remaining cards. Find the probability of getting

(a) neither a heart nor a king

(b) neither an ace nor a king

(c) neither a red card nor a queen card

(d) a black card or an ace.

(e) either a heart or a spade card

(f) a king card

(g) a heart card

(h) a red card

(i) a black card

(j) a spade card

(k) a diamond card

(l) a club card

(m) either an ace card or black card

(n) an ace card

(o) a face card

(p) a face card with red colour

(q) neither ‘10’ card nor an ace

(r) an even number card

(s) an odd number card

(t) not a natural number.

7. Five cards—the ten, jack, queen, king and ace of diamonds, are removed from the well-shuffled 52 playing cards. One card is then picked up at random. Find the probability of getting

(a) neither a heart nor a king

(b) neither an ace nor a king

(c) neither a red card nor a queen card

(d) a black card or an ace.

(e) either a heart or a spade card

(f) a king card

(g) a heart card

(h) a red card

(i) a black card

(j) a spade card

(k) a diamond card

(l) a club card

(m) either an ace card or black card

(n) an ace card

(o) a face card

(p) a face card with red colour

(q) neither ‘10’ card nor an ace

(r) an even number card

(s) an odd number card

(t) not a natural number.

IMPORTANT TIPS

• Coin. A coin has two faces termed as Head and Tail.
• Dice. A dice is a small cube which has between one to six spots or numbers on its sides, which is used in board games like Ludo, Snakes & Ladders. In old fashioned English, dice was used only as a plural form, but now dice is used as both the singular and plural form.
• Cards. A pack of playing cards consists of four suits called Hearts, Spades, Diamonds and Clubs. Each suit consists of 13 cards. Each suit consists of one ace, three face cards and nine number cards.

LEARNING OBJECTIVES

At the end of this lesson you will be able to

• define the sample space accurately.
• find that the total number of favourable outcomes are those outcomes in the sample space that are favourable to the occurence of an event.
• find the probability of an event.

Very Short Answer 

Question. In a family of 3 children calculate the probability of having at least one boy.
Answer: Sol. Let ‘b’ denote a boy and ‘g’ denote a girl. Now, the following outcomes may be noted.
S = {bbb, bbg, bgb, bgg, gbg, ggb, gbb, ggg}
\( \therefore \) Total outcomes = 8
No. of outcomes with atleast one boy = 7
\( \therefore \) Probability of having atleast one boy = \( \frac{7}{8} \)

Question. A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
Answer: Sol. Total English alphabets = 26
Number of consonants = 21
\( \therefore \) P (letter is a consonant) = \( \frac{21}{26} \)

Question. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.
Answer: Sol. S = 52
P(neither a red card nor a queen)
= 1 – P(red card or a queen)
= \( 1 - \left( \frac{26 + 4 - 2}{52} \right) \)
Red cards = 26
Queen = 4
Red queen = 2
= \( 1 - \frac{28}{52} = \frac{24}{52} = \frac{6}{13} \)

Question. A box contains cards numbered 6 to 50. A card is drawn at random from the box. Calculate the probability that the drawn card has a number which is a perfect square.
Answer: Sol. Total number of cards = 50 – 6 + 1 = 45
Perfect square numbers are 9, 16, 25, 36, 49, i.e., 5 numbers
\( \therefore \) P(a prefect square) = \( \frac{5}{45} = \frac{1}{9} \)

Question. Cards marked with number 3, 4, 5, ...., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
Answer: Sol. Total no. of cards = 50 – 3 + 1 = 48
Perfect square number cards are 4, 9, 16, 25, 36, 49 i.e., 6 cards
\( \therefore \) P(perfect square number) = \( \frac{6}{48} = \frac{1}{8} \)

Question. A number is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1?
Answer: Sol. \( (-3)^2 = 9 \); \( (-2)^2 = 4 \); \( (-1)^2 = 1 \); \( (0)^2 = 0 \); \( (1)^2 = 1 \); \( (2)^2 = 4 \); \( (3)^2 = 9 \)
\( \therefore \) P(Sq. of nos. \( \leq \) 1) = \( \frac{3}{7} \)

Question. If two different dice are rolled together, calculate the probability of getting an even number on both dice.
Answer: Sol. Two dice can be thrown as 6 \( \times \) 6 = 36 ways. Even numbers on both the dice can be obtained as {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}, i.e., 9 ways.
\( \therefore \) P(even numbers) = \( \frac{9}{36} = \frac{1}{4} \)

Question. Two different dices are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.
Answer: Sol. Total outcomes = \( 6^n = 6^2 = 36 \)
Possible outcomes having the product of the two numbers on the top of the dice as 6 are (3 \( \times \) 2, 2 \( \times \) 3, 6 \( \times \) 1, 1 \( \times \) 6), i.e, 4.
P(Product of two nos. is 6) = \( \frac{4}{36} = \frac{1}{9} \)

Question. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?
Answer: Sol. P(rotten apples) = \( \frac{\text{No. of rotten apples}}{\text{Total apples}} \)
\( 0.18 = \frac{\text{No. of rotten apples}}{900} \)
\( \therefore \) No. of rotten apples = 900 \( \times \) 0.18 = 162

Short Answer-I 

Question. A coin is tossed two times. Find the probability of getting both heads or both tails.
Answer: Sol. S = {HH, HT, TH, TT} = 4
P (both heads or both tails) = P (both heads) + P (both tails) = \( \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)

Question. In a simultaneous toss of two coins, find the probability of getting: (i) exactly one head, (ii) atmost one head.
Answer: Sol. The sample space is given by S = {HH, HT, TH, TT}. Total events n(S) = 4.
(i) exactly one head = {HT, TH} = 2. P(exactly one head) = \( \frac{2}{4} = \frac{1}{2} \).
(ii) atmost one head = {HT, TH, TT} = 3. P(atmost one head) = \( \frac{3}{4} \).

Question. Two coins are tossed simultaneously. Find the probability of getting at least one head.
Answer: Sol. S = {HH, HT, TH, TT}, i.e., 4.
\( \therefore \) P (atleast one head) = \( \frac{3}{4} \)

Question. Three coins are tossed simultaneously. Find the probability of getting exactly two heads.
Answer: Sol. S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8.
P(exactly two heads) = \( \frac{3}{8} \)

Question. Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.
Answer: Sol. S = {HH, HT, TH, TT}, i.e., 3.
P(at least one tail) = \( \frac{3}{4} \)

Question. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Answer: Sol. P (neither an ace nor a king) = 1 – P (either an ace or a king)
= 1 – [P (an ace) + P (a king)]
= \( 1 - \left[ \frac{4}{52} + \frac{4}{52} \right] \)
= \( 1 - \frac{8}{52} = \frac{44}{52} = \frac{11}{13} \)

Question. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen.
Answer: Sol. P (neither a king nor a queen) = 1 – P (king or queen) = \( 1 - \left( \frac{8}{52} \right) = \frac{11}{13} \)

Question. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a jack nor an ace.
Answer: Sol. Total number of cards = 52. Numbers of jacks = 4. Numbers of aces = 4. Card is neither a jack nor an ace = 52 – 4 – 4 = 44.
\( \therefore \) Required probability = \( \frac{44}{52} = \frac{11}{13} \)

Question. Two dice are thrown simultaneously. Find the probability of getting a doublet.
Answer: Sol. Two dice can be thrown as 6 \( \times \) 6 = 36 ways. “a doublet” can be obtained by (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), i.e., 6 ways.
P(a doublet) = \( \frac{6}{36} = \frac{1}{6} \)

Question. A dice is tossed once. Find the probability of getting an even number or a multiple of 3.
Answer: Sol. S = {1, 2, 3, 4, 5, 6} = 6. ‘an even number or multiple of 3’ are 2, 3, 4, 6, i.e., 4 numbers.
\( \therefore \) Required probability = \( \frac{4}{6} = \frac{2}{3} \)

Question. Two different dice are tossed together. Find the probability (i) that the number on each die is even. (ii) that the sum of numbers appearing on the two dice is 5.
Answer: Sol. Two dice can be thrown as 6 \( \times \) 6 = 36 ways.
(i) The probability of number on each die is even are {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}, i.e., 9 ways. \( \therefore \) P(no. on each die is even) = \( \frac{9}{36} = \frac{1}{4} \).
(ii) “Sum is 5” can be obtained as (2, 3), (3, 2), (1, 4), (4, 1), i.e., 4 ways. \( \therefore \) P(Sum 5) = \( \frac{4}{36} = \frac{1}{9} \)

Short Answer-II 

Question. A coin is tossed two times. Find the probability of getting at least one head.
Answer: Sol. S = {HH, HT, TH, TT}. Total number of ways = 4. Atleast one head = {HH, HT, TH}, i.e., 3 ways. \( \therefore \) P (atleast one head) = \( \frac{3}{4} \)

Question. A coin is tossed two times. Find the probability of getting not more than one head.
Answer: Sol. S = {HH, HT, TH, TT} = 4. Favourable cases are HT, TH, TT i.e., 3 cases. \( \therefore \) P (not more than 1 head) = \( \frac{3}{4} \)

Question. Three distinct coins are tossed together. Find the probability of getting (i) at least 2 heads (ii) at most 2 heads.
Answer: Sol. Total number of possible outcomes = \( 2^n = 2^3 = 8 \) (HHH, TTT, HHT, THH, THT, HTH, TTH, HTT).
(i) Possible outcomes of at least two heads = 4 (HHT, THH, HHH, HTH). \( \therefore \) P(at least two heads) = \( \frac{4}{8} = \frac{1}{2} \).
(ii) Possible outcomes of at most two heads = 7 (HHT, TTT, THH, THT, HTH, TTH, HTT). \( \therefore \) P(at most two heads) = \( \frac{7}{8} \)

Question. Three different coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at least two tails.
Answer: Sol. S {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. Total number of ways = 8.
(i) Exactly two heads = HHT, HTH, THH, i.e., 3 ways. \( \therefore \) P (exactly two heads) = \( \frac{3}{8} \).
(ii) Atleast two heads = HHT, HTH, THH, HHH i.e., 4 ways. \( \therefore \) P (atleast two heads) = \( \frac{4}{8} = \frac{1}{2} \).
(iii) Atleast two tails = TTH, THT, HTT, TTT i.e., 4 ways. \( \therefore \) P (atleast two tails) = \( \frac{4}{8} = \frac{1}{2} \)

Question. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5.
Answer: Sol. Total number of tickets = 40. ‘A multiple of 5’ are 5, 10, 15, 20, … 40, i.e., 8 tickets. \( \therefore \) P (A multiple of 5) = \( \frac{8}{40} = \frac{1}{5} \)

Question. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.
Answer: Sol. Multiples of 3 & 4 are 12, 24, 36, 48, i.e., 4 nos. P(a multiple of 3 and 4) = \( \frac{4}{50} = \frac{2}{25} \)

Question. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is 1/4. The probability of selecting a blue ball at random from the same jar is 1/3. If the jar contains 10 orange balls, find the total number of balls in the jar.
Answer: Sol. P(Red) = \( \frac{1}{4} \), P (blue) = \( \frac{1}{3} \). As we know, Total Probability = 1.
\( \Rightarrow \) P(orange) = \( 1 - \frac{1}{4} - \frac{1}{3} = \frac{5}{12} \).
\( \Rightarrow \text{P(orange)} = \frac{\text{Total no. of orange balls}}{\text{Total no. of balls}} \).
\( \Rightarrow \frac{5}{12} = \frac{10}{\text{Total no. of balls}} \). \( \therefore \) Total no. of balls = \( \frac{10 \times 12}{5} = 24 \).

Question. A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is 3/10 and that of a black ball is 2/5, then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.
Answer: Sol. Let W, B and R denote the White, Black and Red balls respectively. We know, Total probability = 1.
P(W) = \( \frac{3}{10} \), P(B) = \( \frac{2}{5} \). \( \therefore \) P(R) = \( 1 - \frac{3}{10} - \frac{2}{5} = \frac{3}{10} \).
P(B) \( \times \) Total no. of balls = No. of Black balls. \( \frac{2}{5} \times \text{(Total number of balls)} = 20 \).
\( \therefore \) Total number of balls = \( \frac{20 \times 5}{2} = 50 \).

Question. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.
Answer: Sol. Let the number of black balls = x. The number of white balls = 15. \( \therefore \) Total number of balls = x + 15.
\( \therefore \) P(black ball) = 3P(White balls). \( \frac{x}{x + 15} = 3 \left( \frac{15}{x + 15} \right) \). \( \frac{x}{x + 15} = \frac{45}{x + 15} \).
\( \because x + 15 \neq 0, \therefore x = 45 \). Hence, the number of black balls = 45.

Question. In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11?
Answer: Sol. Two dice can be thrown in 6 \( \times \) 6 = 36 ways.
(i) “a prime number on each dice” can be obtained as (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5), i.e., 9 ways.
\( \therefore \) P(a prime no. on each dice) = \( \frac{9}{36} = \frac{1}{4} \).
(ii) “a total of 9 or 11” can be obtained as (3,6),(6,3),(4,5),(5,4) [Total '9'] and (5,6),(6,5) [Total '11'], i.e., 6 ways.
\( \therefore \) P(a total of 9 or 11) = \( \frac{6}{36} = \frac{1}{6} \)

Question. A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal’ another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that: (i) Ramesh will buy the selected shirt? (ii) ‘Kewal’ will buy the selected shirt?
Answer: Sol. Total = 88 + 8 + 4 = 100. No. of good shirts = 88.
(i) P(Ramesh buys a shirt) = P(good shirts) = \( \frac{88}{100} = \frac{22}{25} \).
(ii) No. of shirts without major defect = 96. P(Kewal buys a shirt) = P(shirts without major defect) = \( \frac{88 + 8}{100} = \frac{96}{100} = \frac{24}{25} \)

Question. A card is drawn at random from a well shuffled pack of 52 cards. Find the probability of getting: (i) a red king. (ii) a queen or a jack.
Answer: Sol. Cards in a pack = 52. Number of kings = 4. Number of red kings = 2.
(i) P(a red king) = \( \frac{2}{52} = \frac{1}{26} \).
(ii) P(a queen or a jack) = P(a queen) + P(a jack) = \( \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13} \)

Question. All red face cards are removed from a pack of playing cards. The remaining cards were well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is (i) a red card (ii) a face card (iii) a card of clubs.
Answer: Sol. Number of red face cards removed = 6. \( \therefore \) Remaining cards = 52 – 6 = 46. Hence, Total no. of outcomes = 46.
(i) Possible outcomes of red cards = 26 – 6 = 20. \( \therefore \) P(a red card) = \( \frac{20}{46} = \frac{10}{23} \).
(ii) Possible outcomes of face cards = 6. \( \therefore \) P(a face card) = \( \frac{6}{46} = \frac{3}{23} \).
(iii) Possible outcomes of card of clubs = 13. \( \therefore \) P(a card of clubs) = \( \frac{13}{46} \)

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: Sol. Total cards in the pack = 52. Cards removed = 2(Jacks) + 2(Queens) + 2(Kings) = 6. \( \therefore \) Remaining cards = 52 – 6 = 46.
(i) Number of black Kings = 2. \( \therefore \) P(a black King) = \( \frac{2}{46} = \frac{1}{23} \).
(ii) Total red cards in the pack = 26. Red cards removed = 6. Remaining red cards = 26 – 6 = 20. \( \therefore \) P(a card of red colour) = \( \frac{20}{46} = \frac{10}{23} \).
(iii) Total black cards in the pack = 26. \( \therefore \) P(a card of black colour) = \( \frac{26}{46} = \frac{13}{23} \)

Question. There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the prob ability that the number on the selected card (i) is divisible by 9 and is a perfect square (ii) is a prime number greater than 80.
Answer: Sol. Total cards = 100.
(i) Numbers which are “Divisible by 9 and perfect squares” are 9, 36, 81, i.e., 3. \( \therefore \) P(Divisible by 9 & perfect square) = \( \frac{3}{100} \).
(ii) Numbers which are “prime numbers greater than 80” are 83, 89, 97, i.e., 3. \( \therefore \) P(Prime nos. > 80) = \( \frac{3}{100} \)

Question. Two different dice are rolled together. Find the probability of getting: (i) the sum of numbers on two dice to be 5. (ii) even numbers on both dice.
Answer: Sol. Total possible outcomes = \( 6^n = 6^2 = 36 \).
(i) The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) when the sum of numbers on two dice is 5, i.e., 4. \( \therefore \) Required Probability, P(E) = \( \frac{4}{36} = \frac{1}{9} \).
(ii) The possible outcomes are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) for even numbers on both dice; 9. \( \therefore \) Required Probability, P(E) = \( \frac{9}{36} = \frac{1}{4} \)

Question. Two different dice are thrown together. Find the probability of: (i) getting a number greater than 3 on each die (ii) getting a total of 6 or 7 of the numbers on two dice.
Answer: Sol. Two dice can be thrown in 6 \( \times \) 6 = 36 ways.
(i) “getting a number > 3 on each die” can be obtained as (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6), i.e., 9 ways. P(number > 3 on each die) = \( \frac{9}{36} = \frac{1}{4} \).
(ii) “a total of 6 or 7 can be obtained as (2, 4),(4,2),(3,3),(1,5),(5,1) [Total '6'] and (1,6),(6,1),(2,5),(5,2),(3,4),(4,3) [Total '7'], i.e., 11 ways. P(a total of 6 or 7) = \( \frac{11}{36} \)

Question. Two different dice are thrown together. Find the probability that the numbers obtained. (i) have a sum less than 6 (ii) have a product less than 16 (iii) is a doublet of odd numbers.
Answer: Sol. Two dice can be thrown in (6 \( \times \) 6) = 36 ways.
(i) Sum less than ‘6’ are (1, 1) (1, 2) (1, 3) (1, 4), (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (4, 1) i.e., 10 ways. \( \therefore \) P(sum < 6) = \( \frac{10}{36} = \frac{5}{18} \).
(ii) Product less than ‘16’ 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) i.e., 25 ways. \( \therefore \) P(product less than 16) = \( \frac{25}{36} \).
(iii) Doublet of odd numbers: (1, 1), (3, 3), (5, 5) i.e., 3 ways. \( \therefore \) P(doublet of odd nos.) = \( \frac{3}{36} = \frac{1}{12} \)

Please click the link below to download CBSE Class 10 Probability Sure Shot Questions Set B.