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VOOZH | about |
This exercise set introduces 8th grade students to the fundamental concepts of probability, a crucial component of data handling. It covers the basics of calculating probabilities for simple events, understanding the probability scale from 0 to 1, and applying these concepts to real-world scenarios. Students will encounter problems involving coin tosses, die rolls, card draws, and other everyday situations to develop their intuition about chance and likelihood. The exercises aim to build a strong foundation in probabilistic thinking, preparing students for more complex statistical concepts in future studies. By working through these problems, students will enhance their analytical skills and learn to quantify uncertainty in various situations.
Solution:
Let event of raining tomorrow be P (A)
The probability of raining tomorrow is P (A) = 0.85
Probability of not raining is given by P (A) = 1 – P (A)
Therefore, probability of not raining tomorrow = P (A) = 1 – 0.85
= 0.15
Solution:
Outcomes of a die are 1, 2, 3, 4, 5, 5 and 6
Total outcomes = 6
Prime numbers are 1, 3 and 5
Favorable outcomes = 3
Probability of getting a prime number = Number of favorable outcomes/Total outcomes
= 3/6
= 1/2
Therefore, probability of getting a prime number = 1/2
Solution:
Outcomes of a die are 1, 2, 3, 4, 5, 5 and 6
Total outcomes = 6
Favorable outcomes=2
Probability of getting 2 and 4 = Favorable outcomes/Total outcomes
= 2/6
= 1/3
Therefore, probability of getting 2 and 4 is 1/3
Solution:
Outcomes of a die are 1, 2, 3, 4, 5, 5 and 6
Total outcomes = 6
Multiples of 2 and 3 are 2, 3, 4 and 6
Favorable outcomes = 4
Total number of multiples are 4
Probability of getting a multiple of 2 or 3 = Favorable outcomes/Total number of outcomes
= 4/6
= 2/3
Therefore, probability of getting a multiple of 2 or 3 is 2/3
Solution:
Possible outcomes when a pair of dice is rolled 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), (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)
(where first number shows the number on first dice and second number shows the number on second dice.)
Total number of outcomes = 36
Number of outcomes having 8 as sum are (6, 2), (5, 3), (4, 4), (3, 5) and (2, 6)
Numbers of favorable outcomes = 5
Probability of getting numbers of outcomes having 8 as sum = Favorable outcomes/Total outcomes
= 5/36
Therefore, probability of getting numbers of outcomes having 8 as sum is 5/36
Solution:
Total number of outcomes = 36
Number of outcomes as doublet are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)
Number of favorable outcomes = 6
Probability of getting numbers of outcomes as doublet = Favorable outcomes/Total outcomes
= 6/36
= 1/6
Therefore, probability of getting numbers of outcomes as doublet is 1/6
Solution:
Total number of outcomes = 36
Number of outcomes as doublet of prime numbers are (1, 1), (3, 3), (5, 5)
Number of favorable outcomes = 3
Probability of getting numbers of outcomes as doublet of prime numbers = Favorable outcomes/Total outcomes
= 3/36
= 1/12
Therefore, probability of getting numbers of outcomes as doublet of prime numbers is 1/12
Solution:
Total number of outcomes = 36
Number of outcomes as doublet of odd numbers are (1, 1), (3, 3), (5, 5)
Number of favorable outcomes = 3
Probability of getting numbers of outcomes as doublet of odd numbers = Favorable outcomes/Total outcomes
= 3/36
= 1/12
Therefore, probability of getting numbers of outcomes as doublet of odd numbers is 1/12
Solution:
Total number of outcomes = 36
Number of outcomes having sum greater than 9 are (4, 6), (5, 5), (5, 6), (6, 6), (6, 4), (6, 5)
Number of favorable outcomes = 6
Probability of getting numbers of outcomes having sum greater than 9 = Favorable outcomes/Total outcomes
= 6/36
= 1/6
Therefore, probability of getting numbers of outcomes having sum greater than 9 is 1/6
Solution:
Total number of outcomes = 36
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)
Number of favorable outcomes = 18
Probability of getting numbers of outcomes having an even number on first = Favorable outcomes/Total outcomes
= 18/36
= 1/2
Therefore, probability of getting numbers of outcomes having an even number on first is 1/2
Solution:
Total number of outcomes = 36
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)
Number of favorable outcomes = 6
Probability of getting an even number on one and a multiple of 3 on the other is = Favorable outcomes/Total outcomes
= 6/36
= 1/6
Therefore, probability of getting an even number on one and a multiple of 3 on the other is 1/6
Solution:
Total number of outcomes = 36
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)
Number of favorable outcomes for 9 nor 11 as the sum of the numbers on the faces are 6
Probability of getting 9 nor 11 as the sum of the numbers on the faces is = Favorable outcomes/Total outcomes
= 6/36
= 1/6
Probability of outcomes having 9 nor 11 as the sum of the numbers on the faces P (E) = 1/6
Probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is given by P (E) = 1 – 1/6 = (6 - 1)/5 = 5/6
Therefore, probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is 5/6
Solution:
Total number of outcomes = 36
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)
Number of favorable outcomes = 10
Probability of getting a sum less than 6 is = Favorable outcomes/Total outcomes
= 10/36
= 5/18
Therefore, probability of getting sum less than 6 is 5/18
Solution:
Total number of outcomes = 36
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)
Number of favorable outcomes = 15
Probability of getting a sum less than 7 is = Favorable outcomes/Total outcomes
= 15/36
= 5/12
Therefore, probability of getting sum less than 7 is 5/12
Solution:
Total number of outcomes = 36
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)
Number of favorable outcomes = 15
Probability of getting a sum more than 7 is = Favorable outcomes/Total outcomes
= 15/36
= 5/12
Therefore, probability of getting sum more than 7 is 5/12
Solution:
Total number of outcomes = 36
Number of favorable outcomes =11
Probability of getting outcomes for at least once is = Favorable outcomes/Total outcomes
= 11/36
Therefore, probability of getting outcomes for at least once is 11/36
(xiii) A number other than 5 on any dice.
Solution:
Total number of outcomes = 36
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)
Number of favorable outcomes having 5 on any die = 15
Probability of getting 5 on any die is = Favorable outcomes/Total outcomes
= 11/36
Therefore, probability of getting 5 on any die is 11/36
Probability of not getting 5 on any die P (E) = 1 – P (E)
= 1 – 11/36
= (36 - 11)/36
= 25/36
Therefore, probability of not getting 5 on any die is 25/36
Solution:
Possible outcome of tossing three coins are HTT, HHT, HHH, HTH, TTT, TTH, THT, THH
Total outcomes = 8
Number of outcomes of exactly two heads are HHT,HTH,THH
Favorable outcomes = 3
Probability of getting exactly two heads is = Favorable outcomes/Total outcomes
= 3/8
Therefore, probability of getting exactly two heads is 3/8
Solution:
Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH
Total outcomes = 8
Number of outcomes of at least two heads are HHT,HHH,HTH,THH
Favorable outcomes = 4
Probability of getting at least two heads = Favorable outcomes/Total outcomes
= 4/8
= 1/2
Therefore, probability of getting at least two heads is 1/2
Solution:
Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH
Total outcomes = 8
Number of outcomes of at least one head and one tail are HTT, HHT, HTH, TTH, THT, THH
Favorable outcomes = 6
Probability of getting at least one head and one tail = Favorable outcomes/Total outcomes
= 6/8
= 3/4
Therefore, probability of getting at least one head and one tail is 3/4
Solution:
Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH
Total outcomes = 8
Number of outcomes of no tails are HHH
Favorable outcomes = 1
Probability of getting no tails = Favorable outcomes/Total outcomes
= 1/8
Therefore, probability of getting no tails is 1/8
Solution:
Total number of cards are 52
Number of black king cards are 2
Probability of getting black king cards is = Favorable outcomes/Total outcomes
= 2/52
= 1/26
Therefore, probability of getting black king cards is 1/26
Solution:
Total number of cards are 52
Number of either a black card or a king = 28
Probability of getting either a black card or a king is = Favorable outcomes/Total outcomes
= 28/52
= 7/13
Therefore, probability of getting either a black card or a king is 7/13
Solution:
Total number of cards are 52
Number of black and a king are 2
Probability of getting black and a king is = Favorable outcomes/Total outcomes
= 2/52
= 1/26
Therefore, probability of getting black and a king is 1/26
Solution:
Total number of cards are 52
Number of a jack, queen or a king = 12
Probability of getting a jack, queen or a king is = Favorable outcomes/Total outcomes
= 12/52
= 3/13
Therefore, probability of getting a jack, queen or a king is 3/13
Solution:
Total numbers of cards = 52
Total number of heart cards = 13
Probability of getting a heart is = Favorable outcomes/Total outcomes
= 13/52
= 1/4
Total number of king cards = 4
Probability of getting a king is = Favorable outcomes/Total outcomes
= 4/52
= 1/13
One card is common in heart and king (king of heart)
Total probability of getting a heart and a king = 1/4+ 1/13 – 1/52
= (13 + 4 - 1)/52
= (17 - 1)/52
= 16/52
= 4/13
Therefore, probability of getting neither a heart nor a king = 1 – 4/13 = (13 - 4)/13 = 9/13
Solution:
Total numbers of cards = 52
Number of spade cards = 13
Probability of getting spade cards is = Favorable outcomes/Total outcomes
= 13/52
= 1/4
Number of ace cards = 4
Probability of getting ace cards is = Favorable outcomes/Total outcomes
= 4/52
= 1/13
One card is common in both ace and spade (ace of spade) = 1/52
Probability of getting an ace or spade cards is = 1/4 + 1/13 – 1/52
= (13 + 4 - 1)/52
= (17 - 1)/52
= 16/52
= 4/13
Therefore, probability of getting an ace or spade cards is = 4/13
Solution:
Total numbers of cards = 52
Number of king cards = 4
Number of ace cards = 4
Total number of cards = 4 + 4 = 8
Total number of neither an ace nor a king are = 52 – 8 = 44
Probability of getting neither an ace nor a king is = Favorable outcomes/Total outcomes
= 44/52
= 11/13
Therefore, probability of getting neither an ace nor a king is 11/13
Solution:
Total numbers of cards = 52
Red cards include hearts and diamonds
Number of hearts in a deck of 52 cards = 13
Number of diamonds in a deck of 52 cards = 13
Number of queen in a deck of 52 cards = 4
Total number of red card and queen = 13 + 13 + 2 = 28 (since queen of heart and queen of diamond are already considered)
Number of card which is neither a red card nor a queen = 52 – 28 = 24
Probability of getting neither a king nor a queen is = Favorable outcomes/Total outcomes
= 24/52
= 6/13
Therefore, probability of getting neither a king nor a queen is 6/13
Solution:
Total numbers of cards = 52
Total number of ace cards = 4
Total number of non-ace cards = 52 - 4 = 48
Probability of getting non-ace is = Favorable outcomes/Total outcomes
= 48/52
= 12/13
Therefore, probability of getting non-ace card is 12/13
Solution:
Total numbers of cards are 52
Total number of ten cards = 4
Probability of getting ten cards is = Favorable outcomes/Total outcomes
= 4/52
= 1/13
Therefore, probability of getting ten- card is 1/13
Solution:
Total numbers of cards = 52
Total number of spade cards = 13
Probability of getting spade is = Favorable outcomes/Total outcomes
= 13/52
= 1/4
Therefore, probability of getting a spade is 1/4
Solution:
Total numbers of cards = 52
Cards of spades and clubs are black cards.
Number of spades = 13
Number of clubs = 13
Total number of black card out of 52 cards = 13 + 13 = 26
Probability of getting black cards is = Favorable outcomes/Total outcomes
= 26/52
= 1/2
Therefore, probability of getting a black card is 1/2
Solution:
Total numbers of cards = 52
Total number of the seven of club cards = 1
Probability of getting the seven of clubs cards is = Favorable outcomes/Total outcomes
= 1/52
Therefore, probability of the seven of club card is 1/52
Solution:
Total numbers of cards = 52
Total number of jack cards = 4
Probability of getting jack cards is = Favorable outcomes/Total outcomes
= 4/52
= 1/13
Therefore, probability of the jack card is 1/13
Solution:
Total numbers of cards = 52
Total number of the ace of spades cards = 1
Probability of getting ace of spade cards is = Favorable outcomes/Total outcomes
= 1/52
Therefore, probability of the ace of spade card is 1/52
Solution:
Total numbers of cards = 52
Total number of queen cards = 4
Probability of getting queen cards is = Favorable outcomes/Total outcomes
= 4/52
= 1/13
Therefore, probability of a queen card is 1/13
Solution:
Total numbers of cards = 52
Total number of heart cards = 13
Probability of getting queen cards is = Favorable outcomes/Total outcomes
= 13/52
= 1/4
Therefore, probability of a heart card is 1/4
Solution:
Total numbers of cards = 52
Total number of red cards = 13+13 = 26
Probability of getting queen cards is = Favorable outcomes/Total outcomes
= 26/52
= 1/2
Therefore, probability of a red card is 1/2
Solution:
Total number of red balls = 10
Total number of red white balls = 8
Total number of balls = 10 + 8 = 18
Probability of getting a white ball is = Total number of white balls/Total numbers of balls
= 8/18
= 4/9
Therefore, probability of a white ball is 4/9
Solution:
Total numbers of red balls = 3
Number of black balls = 5
Number of white balls = 4
Total number of balls = 3 + 5 + 4 = 12
Probability of getting a white ball is = Total number of white balls/Total numbers of balls
= 4/12
= 1/3
Therefore, probability of a white ball is 1/3
Solution:
Total numbers of red balls = 3
Number of black balls = 5
Number of white balls = 4
Total number of balls = 3 + 5 + 4 = 12
Probability of getting a red ball is = Total number of red balls/Total numbers of balls
= 3/12
= 1/4
Therefore, probability of a red ball is 1/4
Solution:
Total numbers of red balls = 3
Number of black balls = 5
Number of white balls = 4
Total number of balls = 3 + 5 + 4 = 12
Probability of getting a black ball is = Total number of black balls/Total numbers of balls
= 5/12
Therefore, probability of a black ball is 5/12
Solution:
Total numbers of red balls = 3
Number of black balls = 5
Number of white balls = 4
Total number of Non -red balls = 5 + 4 = 9
Probability of getting a not red ball is = Total number of not red balls/Total numbers of balls
= 9/12
= 3/4
Therefore, probability of not a red ball is 3/4
Solution:
Total numbers are 15
Multiples of 4 are 4, 8, 12
Favorable outcomes=3
Probability of getting a multiple of 4 is = Favorable outcomes/Total outcomes
= 3/15
= 1/5
Therefore, probability of getting multiples of 4 is 1/5
Solution:
Total numbers of red balls = 6
Number of black balls = 8
Number of white balls = 4
Total number of balls = 6 + 8 + 4 = 18
Number of non-black balls are = 6 + 4 = 10
Probability of getting a non-black ball is = Total number of non-black balls/Total number of balls
= 10/18
= 5/9
Therefore, probability of getting a non-black ball is 5/9
Solution:
Total numbers of red balls = 7
Number of white balls = 5
Total number of balls = 7 + 5 = 12
Probability of getting a non-white ball is = Total number of white balls/Total number of balls
= 5/12
Therefore, probability of getting a white ball is 5/12
This exercise set introduces 8th grade students to the fundamental concepts of probability within the broader context of data handling. It covers essential topics such as calculating probabilities for simple events, understanding the probability scale from 0 to 1, and applying these principles to real-world scenarios. Students will encounter a variety of problems involving common probability situations like coin tosses, die rolls, card draws, and marble selections from bags. The exercises are designed to develop students' intuition about chance and likelihood, building a strong foundation for more advanced statistical concepts. By working through these problems, students will enhance their analytical and problem-solving skills, learn to quantify uncertainty, and begin to apply mathematical reasoning to situations involving chance in everyday life.
