Solution:<\/strong><\/p>\n\n\n\n Let event of raining tomorrow be P (A)<\/p>\n\n\n\n The probability of raining tomorrow is P (A) = 0.85<\/p>\n\n\n\n Probability of not raining is given by P (A) = 1 \u2013 P (A)<\/p>\n\n\n\n \u2234 Probability of not raining = P (A) = 1 \u2013 0.85<\/p>\n\n\n\n = 0.15<\/p>\n\n\n\n 2. A die thrown. Find the probability of getting: Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6<\/p>\n\n\n\n Total number of outcome = 6<\/p>\n\n\n\n Prime numbers are: 1, 3 and 5<\/p>\n\n\n\n Total number of prime numbers = 3<\/p>\n\n\n\n Probability of getting a prime number = Total prime numbers\/Total number of outcomes<\/p>\n\n\n\n = 3\/6<\/p>\n\n\n\n = 1\/2<\/p>\n\n\n\n \u2234 Probability of getting a prime number = 1\/2<\/p>\n\n\n\n (ii)<\/strong> Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6<\/p>\n\n\n\n Total number of outcome = 6<\/p>\n\n\n\n Probability of getting 2 and 4 is Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 2\/6<\/p>\n\n\n\n = 1\/3<\/p>\n\n\n\n \u2234 Probability of getting 2 and 4 is 1\/3<\/p>\n\n\n\n (iii)<\/strong> Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6<\/p>\n\n\n\n Multiples of 2 and 3 are = 2, 3, 4 and 6<\/p>\n\n\n\n Total number of multiples are 4<\/p>\n\n\n\n Probability of getting a multiple of 2 or 3 is Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 4\/6<\/p>\n\n\n\n = 2\/3<\/p>\n\n\n\n \u2234 Probability of getting a multiple of 2 or 3 is 2\/3<\/p>\n\n\n\n 3. In a simultaneous throw of a pair of dice, find the probability of getting: Solution:<\/strong><\/p>\n\n\n\n Let us construct a table.<\/p>\n\n\n\n Here the first number denotes the outcome of first die and second number denotes the outcome of second die.<\/p>\n\n\n\n (i)<\/strong> 8 as the sum<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes having 8 as sum are: (6, 2), (5, 3), (4, 4), (3, 5) and (2, 6)<\/p>\n\n\n\n Therefore numbers of outcomes having 8 as sum are 5<\/p>\n\n\n\n Probability of getting numbers of outcomes having 8 as sum is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 5\/36<\/p>\n\n\n\n \u2234 Probability of getting numbers of outcomes having 8 as sum is 5\/36<\/p>\n\n\n\n (ii)<\/strong> a doublet<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes as doublet are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)<\/p>\n\n\n\n Number of outcomes as doublet are 6<\/p>\n\n\n\n Probability of getting numbers of outcomes as doublet is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 6\/36<\/p>\n\n\n\n = 1\/6<\/p>\n\n\n\n \u2234 Probability of getting numbers of outcomes as doublet is 1\/6<\/p>\n\n\n\n (iii)<\/strong> a doublet of prime numbers<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes as doublet of prime numbers are: (1, 1), (3, 3), (5, 5)<\/p>\n\n\n\n Number of outcomes as doublet of prime numbers are 3<\/p>\n\n\n\n Probability of getting numbers of outcomes as doublet of prime numbers is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 3\/36<\/p>\n\n\n\n = 1\/12<\/p>\n\n\n\n \u2234 Probability of getting numbers of outcomes as doublet of prime numbers is 1\/12<\/p>\n\n\n\n (iv)<\/strong> a doublet of odd numbers<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes as doublet of odd numbers are: (1, 1), (3, 3), (5, 5)<\/p>\n\n\n\n Number of outcomes as doublet of odd numbers are 3<\/p>\n\n\n\n Probability of getting numbers of outcomes as doublet of odd numbers is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 3\/36<\/p>\n\n\n\n = 1\/12<\/p>\n\n\n\n \u2234 Probability of getting numbers of outcomes as doublet of odd numbers is 1\/12<\/p>\n\n\n\n (v)<\/strong> a sum greater than 9<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes having sum greater than 9 are: (4, 6), (5, 5), (5, 6), (6, 6), (6, 4), (6, 5)<\/p>\n\n\n\n Number of outcomes having sum greater than 9 are 6<\/p>\n\n\n\n Probability of getting numbers of outcomes having sum greater than 9 is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 6\/36 = 1\/6<\/p>\n\n\n\n \u2234 Probability of getting numbers of outcomes having sum greater than 9 is 1\/6<\/p>\n\n\n\n (vi)<\/strong> An even number on first<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes having an even number on first are: (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) and (6, 6)<\/p>\n\n\n\n Number of outcomes having an even number on first are 18<\/p>\n\n\n\n Probability of getting numbers of outcomes having an even number on first is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 18\/36<\/p>\n\n\n\n = 1\/2<\/p>\n\n\n\n \u2234 Probability of getting numbers of outcomes having an even number on first is 1\/2<\/p>\n\n\n\n (vii)<\/strong> An even number on one and a multiple of 3 on the other<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes having an even number on one and a multiple of 3 on the other are: (2, 3), (2, 6), (4, 3), (4, 6), (6, 3) and (6, 6)<\/p>\n\n\n\n Number of outcomes having an even number on one and a multiple of 3 on the other are 6<\/p>\n\n\n\n Probability of getting an even number on one and a multiple of 3 on the other is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 6\/36<\/p>\n\n\n\n = 1\/6<\/p>\n\n\n\n \u2234 Probability of getting an even number on one and a multiple of 3 on the other is 1\/6<\/p>\n\n\n\n (viii)<\/strong> Neither 9 nor 11 as the sum of the numbers on the faces<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes having 9 nor 11 as the sum of the numbers on the faces are: (3, 6), (4, 5), (5, 4), (5, 6), (6, 3) and (6, 5)<\/p>\n\n\n\n Number of outcomes having neither 9 nor 11 as the sum of the numbers on the faces are 6<\/p>\n\n\n\n Probability of getting 9 nor 11 as the sum of the numbers on the faces is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 6\/36<\/p>\n\n\n\n = 1\/6<\/p>\n\n\n\n Probability of outcomes having 9 nor 11 as the sum of the numbers on the faces P (E) = 1\/6<\/p>\n\n\n\n \u2234Probability of getting neither 9 nor 11 as the sum of the numbers on the faces is 1\/6<\/p>\n\n\n\n Probability of outcomes not<\/strong> having 9 nor 11 as the sum of the numbers on the faces is given by P (E) = 1 \u2013 1\/6 = (6-1)\/5 = 5\/6<\/p>\n\n\n\n \u2234 Probability of outcomes not<\/strong> having 9 nor 11 as the sum of the numbers on the faces is 5\/6<\/p>\n\n\n\n (ix)<\/strong> A sum less than 6<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes having a 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)<\/p>\n\n\n\n Number of outcomes having a sum less than 6 are 10<\/p>\n\n\n\n Probability of getting a sum less than 6 is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 10\/36<\/p>\n\n\n\n = 5\/18<\/p>\n\n\n\n \u2234 Probability of getting sum less than 6 is 5\/18<\/p>\n\n\n\n (x)<\/strong> A sum less than 7<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes having a sum less than 7 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, 2), (5, 1)<\/p>\n\n\n\n Number of outcomes having a sum less than 7 are 15<\/p>\n\n\n\n Probability of getting a sum less than 7 is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 15\/36<\/p>\n\n\n\n = 5\/12<\/p>\n\n\n\n \u2234 Probability of getting sum less than 7 is 5\/12<\/p>\n\n\n\n (xi)<\/strong> A sum more than 7<\/p>\n\n\n\n Total number of outcomes in the above table are 36<\/p>\n\n\n\n Number of outcomes having a sum more than 7 are: (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)<\/p>\n\n\n\n Number of outcomes having a sum more than 7 are 15<\/p>\n\n\n\n Probability of getting a sum more than 7 is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 15\/36<\/p>\n\n\n\n = 5\/12<\/p>\n\n\n\n \u2234 Probability of getting sum more than 7 is 5\/12<\/p>\n\n\n\n (xii)<\/strong> At least once<\/p>\n\n\n\n Total number of outcomes in the above table 1 are 36<\/p>\n\n\n\n Number of outcomes for at least once are 11<\/p>\n\n\n\n Probability of getting outcomes for at least once is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 11\/36<\/p>\n\n\n\n \u2234 Probability of getting outcomes for at least once is 11\/36<\/p>\n\n\n\n (xiii)<\/strong> A number other than 5 on any dice.<\/p>\n\n\n\n Total number of outcomes in the above table 1 are 36<\/p>\n\n\n\n Number of outcomes having 5 on any die are: (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)<\/p>\n\n\n\n Number of outcomes having outcomes having 5 on any die are 15<\/p>\n\n\n\n Probability of getting 5 on any die is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 11\/36<\/p>\n\n\n\n \u2234 Probability of getting 5 on any die is 11\/36<\/p>\n\n\n\n Probability of not<\/strong> getting 5 on any die P (E) = 1 \u2013 P (E)<\/p>\n\n\n\n = 1 \u2013 11\/36<\/p>\n\n\n\n = (36-11)\/36<\/p>\n\n\n\n = 25\/36<\/p>\n\n\n\n 4. Three coins are tossed together. Find the probability of getting: Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> Exactly two heads<\/p>\n\n\n\n Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH<\/p>\n\n\n\n Number of outcomes of exactly two heads are: 3<\/p>\n\n\n\n Probability of getting exactly two heads is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 3\/8<\/p>\n\n\n\n \u2234 Probability of getting exactly two heads is 3\/8<\/p>\n\n\n\n (ii)<\/strong> At least two heads<\/p>\n\n\n\n Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH<\/p>\n\n\n\n Number of outcomes of at least two heads are: 4<\/p>\n\n\n\n Probability of getting at least two heads is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 4\/8<\/p>\n\n\n\n = 1\/2<\/p>\n\n\n\n \u2234 Probability of getting at least two heads is 1\/2<\/p>\n\n\n\n (iii)<\/strong> At least one head and one tail<\/p>\n\n\n\n Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH<\/p>\n\n\n\n Number of outcomes of at least one head and one tail are: 6<\/p>\n\n\n\n Probability of getting at least one head and one tail is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 6\/8<\/p>\n\n\n\n = 3\/4<\/p>\n\n\n\n \u2234 Probability of getting at least one head and one tail is 3\/4<\/p>\n\n\n\n (iv)<\/strong> No tails<\/p>\n\n\n\n Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH<\/p>\n\n\n\n Number of outcomes of no tails are: 1<\/p>\n\n\n\n Probability of getting no tails is = Total numbers\/Total number of outcomes<\/p>\n\n\n\n = 1\/8<\/p>\n\n\n\n \u2234 Probability of getting no tails is 1\/8<\/p>\n\n\n\n 5.<\/strong> A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is: Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> A black king<\/p>\n\n\n\n Total number of cards are 52<\/p>\n\n\n\n Number of black king cards = 2<\/p>\n\n\n\n Probability of getting black king cards is = Total number of black king cards\/Total number of cards<\/p>\n\n\n\n = 2\/52<\/p>\n\n\n\n = 1\/26<\/p>\n\n\n\n \u2234 Probability of getting black king cards is 1\/26<\/p>\n\n\n\n (ii)<\/strong> Either a black card or a king<\/p>\n\n\n\n Total number of cards are 52<\/p>\n\n\n\n Number of either a black card or a king = 28<\/p>\n\n\n\n Probability of getting either a black card or a king is = Total number of either black or king card\/Total number of cards<\/p>\n\n\n\n = 28\/52<\/p>\n\n\n\n = 7\/13<\/p>\n\n\n\n \u2234 Probability of getting either a black card or a king is 7\/13<\/p>\n\n\n\n (iii)<\/strong> Black and a king<\/p>\n\n\n\n Total number of cards are 52<\/p>\n\n\n\n Number of black and a king = 2<\/p>\n\n\n\n Probability of getting black and a king is = Total number of black and king card\/Total number of cards<\/p>\n\n\n\n = 2\/52<\/p>\n\n\n\n = 1\/26<\/p>\n\n\n\n \u2234 Probability of getting black and a king is 1\/26<\/p>\n\n\n\n (iv)<\/strong> a jack, queen or a king<\/p>\n\n\n\n Total number of cards are 52<\/p>\n\n\n\n Number of a jack, queen or a king = 12<\/p>\n\n\n\n Probability of getting a jack, queen or a king is = Total number of jack, queen or king card\/Total number of cards<\/p>\n\n\n\n = 12\/52<\/p>\n\n\n\n = 3\/13<\/p>\n\n\n\n \u2234 Probability of getting a jack, queen or a king is 3\/13<\/p>\n\n\n\n (v)<\/strong> Neither a heart nor a king<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of heart cards = 13<\/p>\n\n\n\n Probability of getting a heart is = Total number of hearts\/Total number of cards<\/p>\n\n\n\n = 13\/52<\/p>\n\n\n\n = 1\/4<\/p>\n\n\n\n Total number of king cards = 4<\/p>\n\n\n\n Probability of getting a king is = Total number of king card\/Total number of cards<\/p>\n\n\n\n = 4\/52<\/p>\n\n\n\n = 1\/13<\/p>\n\n\n\n Total probability of getting a heart and a king = 13\/52 + 4\/52 \u2013 1\/52<\/p>\n\n\n\n = (13+4-1)\/52<\/p>\n\n\n\n = 16\/52<\/p>\n\n\n\n = 4\/13<\/p>\n\n\n\n \u2234 Probability of getting neither a heart nor a king = 1 \u2013 4\/13 = (13-4)\/13 = 9\/13<\/p>\n\n\n\n (vi)<\/strong> Spade or an ace<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Number of spade cards = 13<\/p>\n\n\n\n Probability of getting spade cards is = Total number of spade card\/Total number of cards<\/p>\n\n\n\n = 13\/52<\/p>\n\n\n\n Number of ace cards = 4<\/p>\n\n\n\n Probability of getting ace cards is = Total number of ace card\/Total number of cards<\/p>\n\n\n\n = 4\/52<\/p>\n\n\n\n = 1\/13<\/p>\n\n\n\n Probability of getting ace and spade cards is = Total number of ace and spade card\/Total number of cards<\/p>\n\n\n\n = 1\/52<\/p>\n\n\n\n Probability of getting an ace or spade cards is = 13\/52 + 4\/52 \u2013 1\/52<\/p>\n\n\n\n = (13+4-1)\/52<\/p>\n\n\n\n = 16\/52<\/p>\n\n\n\n = 4\/13<\/p>\n\n\n\n \u2234 Probability of getting an ace or spade cards is = 4\/13<\/p>\n\n\n\n (vii)<\/strong> Neither an ace nor a king<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Number of king cards = 4<\/p>\n\n\n\n Number of ace cards = 4<\/p>\n\n\n\n Total number of cards = 4 + 4 = 8<\/p>\n\n\n\n Total number of neither an ace nor a king are = 52 \u2013 8 = 44<\/p>\n\n\n\n Probability of getting neither an ace nor a king is = Total number of neither ace nor king card\/Total number of cards<\/p>\n\n\n\n = 44\/52<\/p>\n\n\n\n = 11\/13<\/p>\n\n\n\n \u2234 Probability of getting neither an ace nor a king is 11\/13<\/p>\n\n\n\n (viii)<\/strong> Neither a red card nor a queen<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Red cards include hearts and diamonds<\/p>\n\n\n\n Number of hearts in a deck of 52 cards = 13<\/p>\n\n\n\n Number of diamonds in a deck of 52 cards = 13<\/p>\n\n\n\n Number of queen in a deck of 52 cards = 4<\/p>\n\n\n\n Total number of red card and queen = 13 + 13 + 2 = 28 [since queen of heart and queen of diamond are removed]<\/p>\n\n\n\n Number of card which is neither a red card nor a queen = 52 \u2013 28 = 24<\/p>\n\n\n\n Probability of getting neither a king nor a queen is = Total number of neither red nor queen card\/Total number of cards<\/p>\n\n\n\n = 24\/52<\/p>\n\n\n\n = 6\/13<\/p>\n\n\n\n \u2234 Probability of getting neither a king nor a queen is 6\/13<\/p>\n\n\n\n (ix)<\/strong> Other than an ace<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of ace cards = 4<\/p>\n\n\n\n Total number of non-ace cards = 52-4 = 48<\/p>\n\n\n\n Probability of getting non-ace is = Total number of non-ace cards\/Total number of cards<\/p>\n\n\n\n = 48\/52<\/p>\n\n\n\n = 12\/13<\/p>\n\n\n\n \u2234 Probability of getting non-ace card is 12\/13<\/p>\n\n\n\n (x)<\/strong> A ten<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of ten cards = 4<\/p>\n\n\n\n Probability of getting ten cards is = Total number of ten cards\/Total number of cards<\/p>\n\n\n\n = 4\/52<\/p>\n\n\n\n = 1\/13<\/p>\n\n\n\n \u2234 Probability of getting ten card is 1\/13<\/p>\n\n\n\n (xi)<\/strong> A spade<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of spade cards = 13<\/p>\n\n\n\n Probability of getting spade is = Total number of spade cards\/Total number of cards<\/p>\n\n\n\n = 13\/52<\/p>\n\n\n\n = 1\/4<\/p>\n\n\n\n \u2234 Probability of getting a spade is 1\/4<\/p>\n\n\n\n <\/strong> Total numbers of cards are 52<\/p>\n\n\n\n Cards of spades and clubs are black cards.<\/p>\n\n\n\n Number of spades = 13<\/p>\n\n\n\n Number of clubs = 13<\/p>\n\n\n\n Total number of black card out of 52 cards = 13 + 13 = 26<\/p>\n\n\n\n Probability of getting black cards is = Total number of black cards\/Total number of cards<\/p>\n\n\n\n = 26\/52<\/p>\n\n\n\n = 1\/2<\/p>\n\n\n\n \u2234 Probability of getting a black card is 1\/2<\/p>\n\n\n\n (xiii)<\/strong> The seven of clubs<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of the seven of clubs cards = 1<\/p>\n\n\n\n Probability of getting the seven of clubs cards is = Total number of the seven of club cards\/ Total numbers of cards<\/p>\n\n\n\n = 1\/52<\/p>\n\n\n\n \u2234 Probability of the seven of club card is 1\/52<\/p>\n\n\n\n (xiv)<\/strong> Jack<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of jack cards = 4<\/p>\n\n\n\n Probability of getting jack cards is = Total number of jack cards\/ Total numbers of cards<\/p>\n\n\n\n = 4\/52<\/p>\n\n\n\n = 1\/13<\/p>\n\n\n\n \u2234 Probability of the jack card is 1\/13<\/p>\n\n\n\n (xv)<\/strong> The ace of spades<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of the ace of spades cards = 1<\/p>\n\n\n\n Probability of getting ace of spade cards is = Total number of ace of spade cards\/ Total numbers of cards<\/p>\n\n\n\n = 1\/52<\/p>\n\n\n\n \u2234 Probability of the ace of spade card is 1\/52<\/p>\n\n\n\n <\/strong> Total numbers of cards are 52<\/p>\n\n\n\n Total number of queen cards = 4<\/p>\n\n\n\n Probability of getting queen cards is = Total number of queen cards\/Total numbers of cards<\/p>\n\n\n\n = 4\/52<\/p>\n\n\n\n = 1\/13<\/p>\n\n\n\n \u2234 Probability of a queen card is 1\/13<\/p>\n\n\n\n (xvii)<\/strong> A heart<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of heart cards = 13<\/p>\n\n\n\n Probability of getting queen cards is = Total number of heart cards\/Total numbers of cards<\/p>\n\n\n\n = 13\/52<\/p>\n\n\n\n = 1\/4<\/p>\n\n\n\n \u2234 Probability of a heart card is 1\/4<\/p>\n\n\n\n (xviii)<\/strong> A red card<\/p>\n\n\n\n Total numbers of cards are 52<\/p>\n\n\n\n Total number of red cards = 13+13 = 26<\/p>\n\n\n\n Probability of getting queen cards is = Total number of red cards\/Total numbers of cards<\/p>\n\n\n\n = 26\/52<\/p>\n\n\n\n = 1\/2<\/p>\n\n\n\n \u2234 Probability of a red card is 1\/2<\/p>\n\n\n\n 6. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Total number of red balls = 10<\/p>\n\n\n\n Total number of red white balls = 8<\/p>\n\n\n\n Total number of balls = 10 + 8 = 18<\/p>\n\n\n\n Probability of getting a white ball is = Total number of white balls\/Total numbers of balls<\/p>\n\n\n\n = 8\/18<\/p>\n\n\n\n = 4\/9<\/p>\n\n\n\n \u2234 Probability of a white ball is 4\/9<\/p>\n\n\n\n 7. 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 : Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> White?<\/p>\n\n\n\n Total numbers of red balls = 3<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Number of white balls = 4<\/p>\n\n\n\n Total number of balls = 3 + 5 + 4 = 12<\/p>\n\n\n\n Probability of getting a white ball is = Total number of white balls\/Total numbers of balls<\/p>\n\n\n\n = 4\/12<\/p>\n\n\n\n = 1\/3<\/p>\n\n\n\n \u2234 Probability of a white ball is 1\/3<\/p>\n\n\n\n (ii)<\/strong> red?<\/p>\n\n\n\n Total numbers of red balls = 3<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Number of white balls = 4<\/p>\n\n\n\n Total number of balls = 3 + 5 + 4 = 12<\/p>\n\n\n\n Probability of getting a red ball is = Total number of red balls\/Total numbers of balls<\/p>\n\n\n\n = 3\/12<\/p>\n\n\n\n = 1\/4<\/p>\n\n\n\n \u2234 Probability of a red ball is 1\/4<\/p>\n\n\n\n (iii)<\/strong> black?<\/p>\n\n\n\n Total numbers of red balls = 3<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Number of white balls = 4<\/p>\n\n\n\n Total number of balls = 3 + 5 + 4 = 12<\/p>\n\n\n\n Probability of getting a black ball is = Total number of black balls\/Total numbers of balls<\/p>\n\n\n\n = 5\/12<\/p>\n\n\n\n \u2234 Probability of a black ball is 5\/12<\/p>\n\n\n\n (iv)<\/strong> not red?<\/p>\n\n\n\n Total numbers of red balls = 3<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Number of white balls = 4<\/p>\n\n\n\n Total number of Non red balls = 5 + 4 = 9<\/p>\n\n\n\n Probability of getting a not red ball is = Total number of not red balls\/Total numbers of balls<\/p>\n\n\n\n = 9\/12<\/p>\n\n\n\n = 3\/4<\/p>\n\n\n\n \u2234 Probability of not a red ball is 3\/4<\/p>\n\n\n\n 8. What is the probability that a number selected from the numbers 1, 2, 3, \u2026, 15 is a multiple of 4?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Total numbers are 15<\/p>\n\n\n\n Multiples of 4 are = 4, 8, 12<\/p>\n\n\n\n Probability of getting a multiple of 4 is = Total number of multiples of 4\/Total numbers<\/p>\n\n\n\n = 3\/15<\/p>\n\n\n\n = 1\/5<\/p>\n\n\n\n \u2234 Probability of getting multiples of 4 is 1\/5<\/p>\n\n\n\n 9. A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Total numbers of red balls = 6<\/p>\n\n\n\n Number of black balls = 8<\/p>\n\n\n\n Number of white balls = 4<\/p>\n\n\n\n Total number of non-red balls = 6 + 8 + 4 = 18<\/p>\n\n\n\n Number of non-black balls are = 6 + 4 = 10<\/p>\n\n\n\n Probability of getting a non-black ball is = Total number of non-black balls\/Total number of balls<\/p>\n\n\n\n = 10\/18<\/p>\n\n\n\n = 5\/9<\/p>\n\n\n\n \u2234 Probability of getting a non-black ball is 5\/9<\/p>\n\n\n\n 10. A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Total numbers of red balls = 7<\/p>\n\n\n\n Number of white balls = 5<\/p>\n\n\n\n Total number of Non red balls = 7 + 5 = 12<\/p>\n\n\n\n Probability of getting a non-white ball is = Total number of non-white balls\/Total number of balls<\/p>\n\n\n\n = 5\/12<\/p>\n\n\n\n \u2234 Probability of getting a non-white ball is 5\/12<\/p>\n\n\n\n 11. A bag contains 4 red, 5 black and 6 white balls. One ball is drawn from the bag at random. Find the probability that the ball drawn is: Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> White<\/p>\n\n\n\n Total numbers of red balls = 4<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Number of white balls = 6<\/p>\n\n\n\n Total number of balls = 4 + 5 + 6 = 15<\/p>\n\n\n\n Probability of getting a white ball is = Total number of white balls\/Total number of balls<\/p>\n\n\n\n = 6\/15<\/p>\n\n\n\n = 2\/5<\/p>\n\n\n\n \u2234 Probability of getting a white ball is 2\/5<\/p>\n\n\n\n (ii)<\/strong> Red<\/p>\n\n\n\n Total numbers of red balls = 4<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Number of white balls = 6<\/p>\n\n\n\n Total number of balls = 4 + 5 + 6 = 15<\/p>\n\n\n\n Probability of getting a red ball is = Total number of red balls\/Total number of balls<\/p>\n\n\n\n = 4\/15<\/p>\n\n\n\n \u2234 Probability of getting a white ball is 4\/15<\/p>\n\n\n\n (iii)<\/strong> Not black<\/p>\n\n\n\n Total numbers of red balls = 4<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Number of white balls = 6<\/p>\n\n\n\n Total number of balls = 4 + 5 + 6 = 15<\/p>\n\n\n\n Number of non-black balls = 4 + 6 = 10<\/p>\n\n\n\n Probability of getting a non-black ball is = Total number of non-black balls\/Total number of balls<\/p>\n\n\n\n = 10\/15<\/p>\n\n\n\n = 2\/3<\/p>\n\n\n\n \u2234 Probability of getting a non-black ball is 2\/3<\/p>\n\n\n\n (iv)<\/strong> Red or white<\/p>\n\n\n\n Total numbers of red balls = 4<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Number of white balls = 6<\/p>\n\n\n\n Total number of balls = 4 + 5 + 6 = 15<\/p>\n\n\n\n Number of red and white balls = 4 + 6 = 10<\/p>\n\n\n\n Probability of getting a red or white ball is = Total number of red or white balls\/Total number of balls<\/p>\n\n\n\n = 10\/15<\/p>\n\n\n\n = 2\/3<\/p>\n\n\n\n \u2234 Probability of getting a red or white ball is 2\/3<\/p>\n\n\n\n 12. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> Red<\/p>\n\n\n\n Total numbers of red balls = 3<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Total number of balls = 3 + 5 = 8<\/p>\n\n\n\n Probability of getting a red ball is = Total number of red balls\/Total number of balls<\/p>\n\n\n\n = 3\/8<\/p>\n\n\n\n \u2234 Probability of getting a red ball is 3\/8<\/p>\n\n\n\n (ii)<\/strong> Black<\/p>\n\n\n\n Total numbers of red balls = 3<\/p>\n\n\n\n Number of black balls = 5<\/p>\n\n\n\n Total number of balls = 3 + 5 = 8<\/p>\n\n\n\n Probability of getting a black ball is = Total number of black balls\/Total number of balls<\/p>\n\n\n\n = 5\/8<\/p>\n\n\n\n \u2234 Probability of getting a black ball is 5\/8<\/p>\n\n\n\n 13. A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> Red<\/p>\n\n\n\n Total numbers of red marbles = 5<\/p>\n\n\n\n Number of white marbles = 8<\/p>\n\n\n\n Number of green marbles = 4<\/p>\n\n\n\n Total number of marbles = 5 + 8 + 4 = 17<\/p>\n\n\n\n Probability of getting a red marble is = Total number of red marbles\/Total number of marbles<\/p>\n\n\n\n = 5\/17<\/p>\n\n\n\n \u2234 Probability of getting a red marble is 5\/17<\/p>\n\n\n\n (ii)<\/strong> White<\/p>\n\n\n\n Total numbers of red marbles = 5<\/p>\n\n\n\n Number of white marbles = 8<\/p>\n\n\n\n Number of green marbles = 4<\/p>\n\n\n\n Total number of marbles = 5 + 8 + 4 = 17<\/p>\n\n\n\n Probability of getting a white marble is = Total number of white marbles\/Total number of marbles<\/p>\n\n\n\n = 8\/17<\/p>\n\n\n\n \u2234 Probability of getting a white marble is 8\/17<\/p>\n\n\n\n (iii)<\/strong> Not green<\/p>\n\n\n\n Total numbers of red marbles = 5<\/p>\n\n\n\n Number of white marbles = 8<\/p>\n\n\n\n Number of green marbles = 4<\/p>\n\n\n\n Total number of marbles = 5 + 8 + 4 = 17<\/p>\n\n\n\n Total number of non-green marbles = 5 + 8 = 13<\/p>\n\n\n\n Probability of getting a non-green marble is = Total number of non-green marbles\/Total number of marbles<\/p>\n\n\n\n = 13\/17<\/p>\n\n\n\n \u2234 Probability of getting a non-green marble is 13\/17<\/p>\n\n\n\n 14. If you put 21 consonants and 5 vowels in a bag. What would carry greater probability? Getting a consonant or a vowel? Find each probability?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Total numbers of consonants = 21<\/p>\n\n\n\n Number of white vowels = 5<\/p>\n\n\n\n Total number of alphabets = 21 + 5 = 26<\/p>\n\n\n\n Probability of getting a consonant is = Total number of consonants\/Total number of alphabets<\/p>\n\n\n\n = 21\/26<\/p>\n\n\n\n Probability of getting a vowel is = Total number of vowels\/Total number of alphabets<\/p>\n\n\n\n = 5\/26<\/p>\n\n\n\n \u2234 The probability of getting a consonant is more.<\/p>\n\n\n\n 15. If we have 15 boys and 5 girls in a class which carries a higher probability? Getting a copy belonging to a boy or a girl. Can you give it a value?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Total numbers of boys in a class = 15<\/p>\n\n\n\n Number of girls in a class = 5<\/p>\n\n\n\n Total number of students = 15 + 5 = 20<\/p>\n\n\n\n Probability of getting a copy of a boy is = Total number of boys\/Total number of students<\/p>\n\n\n\n = 15\/20<\/p>\n\n\n\n = 3\/4<\/p>\n\n\n\n Probability of getting a copy of a girl is = Total number of girls\/Total number of students<\/p>\n\n\n\n = 5\/20<\/p>\n\n\n\n = 1\/4<\/p>\n\n\n\n \u2234 The probability of getting a copy of a boy is more.<\/p>\n\n\n\n 16. It you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is (i) white? (ii) black?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Total numbers of white socks = 6 pairs<\/p>\n\n\n\n Total numbers of black socks = 3 pairs<\/p>\n\n\n\n Total number pairs of socks = 6 + 3 = 9<\/p>\n\n\n\n (i) Probability of getting a white sock is = Total number of white socks\/Total number of socks<\/p>\n\n\n\n = 6\/9<\/p>\n\n\n\n = 2\/3<\/p>\n\n\n\n \u2234 The probability of white socks is 2\/3.<\/p>\n\n\n\n (ii) Probability of getting a black sock is = Total number of black socks\/Total number of socks<\/p>\n\n\n\n = 3\/9<\/p>\n\n\n\n = 1\/3<\/p>\n\n\n\n \u2234 The probability of black socks is 1\/3.<\/p>\n\n\n\n 17. If you have a spinning wheel with 3-green sectors, 1-blue sector and 1-red sector. What is the probability of getting a green sector? Is it the maximum?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Total numbers of green sectors = 3<\/p>\n\n\n\n Total numbers of blue sector = 1<\/p>\n\n\n\n Total numbers of red sector = 1<\/p>\n\n\n\n Total number of sectors = 3 + 1 + 1 = 5<\/p>\n\n\n\n Probability of getting a green sector is = Total number of green sectors\/Total number of sectors<\/p>\n\n\n\n = 3\/5<\/p>\n\n\n\n Probability of getting a blue sector is = Total number of blue sectors\/Total number of sectors<\/p>\n\n\n\n = 1\/5<\/p>\n\n\n\n Probability of getting a red sector is = Total number of red sectors\/Total number of sectors<\/p>\n\n\n\n = 1\/5<\/p>\n\n\n\n Yes, the probability of getting a green sector is maximum.<\/p>\n\n\n\n 18. When two dice are rolled: Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> List the outcomes for the event that the total is odd.<\/p>\n\n\n\n Possible outcomes of two dice are:<\/p>\n\n\n\n Outcomes for the event that the total is odd are: (2, 1), (4, 1), (6, 1), (1, 2), (3, 2), (5, 2), (2, 3), (4, 3), (6, 3), (1, 4), (3, 4), (5, 4), (2, 5), (4, 5), (6, 5), (1, 6), (3, 6), (5, 6)<\/p>\n\n\n\n (ii)<\/strong> Find probability of getting an odd total.<\/p>\n\n\n\n Total numbers of outcomes from two dice are 36<\/p>\n\n\n\n From above table we get that the total number of outcomes for the event of getting an odd total is 18.<\/p>\n\n\n\n Probability of getting an event that the total is odd = Total number of events with odd total\/Total number of events<\/p>\n\n\n\n = 18\/36<\/p>\n\n\n\n = 1\/2<\/p>\n\n\n\n \u2234 The probability of getting an odd total is 1\/2<\/p>\n\n\n\n (iii)<\/strong> List the outcomes for the event that total is less than 5.<\/p>\n\n\n\n Total numbers of outcomes from two dice are 36<\/p>\n\n\n\n Total number of outcomes of the events that total is less than 5 are: (1, 1), (2, 1), (3, 1), (1, 2), (2, 2) and (1, 3)<\/p>\n\n\n\n (iv)<\/strong> Find the probability of getting a total less than 5?<\/p>\n\n\n\n Total numbers of outcomes from two dice are 36<\/p>\n\n\n\n Total number of events that total is less than 5 are: (1, 1), (2, 1), (3, 1), (1, 2), (2, 2) and (1, 3)<\/p>\n\n\n\n Probability of getting an event that total is less than 5 = Total number of events with total less than 5 \/Total number of events<\/p>\n\n\n\n = 6\/36<\/p>\n\n\n\n = 1\/6<\/p>\n\n\n\n \u2234 The probability of getting an event that total is less than 5 is 1\/6<\/p>\n\n\n\n RD Sharma Solutions for Class 8 Maths Chapter 26\u2013Data Handling – IV (Probability)<\/p>\n\n\n\n Download PDF<\/strong>: RD Sharma Solutions for Class 8 Maths Chapter 26\u2013Data Handling – IV (Probability) PDF<\/a><\/p>\n\n\n\n
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3<\/strong><\/p>\n\n\n\n
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) An even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/rd-sharma-solutions-for-class-8-maths-chapter-26.png\" "\\"RD")
<\/figure>\n\n\n\n
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail
(iv) no tails<\/strong><\/p>\n\n\n\n
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) neither a red card nor a queen
(ix) other than an ace
(x) a ten
(xi) a spade
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(xv) the ace of spades
(xvi) a queen
(xvii) a heart
(xviii) a red card<\/strong><\/p>\n\n\n\n
(xii)<\/strong> A black card<\/p>\n\n\n\n
(xvi)<\/strong> A queen<\/p>\n\n\n\n
(i) White?
(ii) red?
(iii) black?
(iv) not red?<\/strong><\/p>\n\n\n\n
(i) white
(ii) red
(iii) not black
(iv) red or white<\/strong><\/p>\n\n\n\n
(i) red
(ii) black<\/strong><\/p>\n\n\n\n
(i) red
(ii) white
(iii) not green<\/strong><\/p>\n\n\n\n
(i) List the outcomes for the event that the total is odd.
(ii) Find probability of getting an odd total.
(iii) List the outcomes for the event that total is less than 5.
(iv) Find the probability of getting a total less than 5?<\/strong><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/rd-sharma-solutions-for-class-8-maths-chapter-26-1.png\" "\\"RD")
<\/figure>\n\n\n\nRD Sharma Solutions for Class 8 Maths Chapter 26: Download PDF<\/strong><\/h2>\n\n\n\n
Chapterwise RD Sharma Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n