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Jamb Mathematics - Lesson Notes on Probability for UTME Candidate

Feb 16 2025 04:06 PM

Osason

Jamb Updates

Probability | Jamb Mathematics

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As you embark on your journey to master probability, may wisdom and clarity illuminate your path, guiding you through each concept with ease. Embrace the beauty of chance and certainty, for understanding these mysteries will empower you to see patterns where others see randomness. May your preparation be fruitful, and may confidence and success accompany you in your exam.
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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Probability? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Probability together and move one step closer to achieving your exam success! Blissful learning.
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Calculation problem on experimental probability (tossing of coin, throwing of a dice etc);

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1. A coin is tossed 100 times, and heads appear 58 times. What is the experimental probability of getting heads?
Solution:
Total number of trials = 100100
Number of times heads appear = 5858
Experimental probability of getting heads = Number of headsTotal trials=58100=0.58\frac{\text{Number of heads}}{\text{Total trials}} = \frac{58}{100} = 0.58

2. A die is rolled 200 times, and the number 4 appears 35 times. What is the experimental probability of rolling a 4?
Solution:
Total number of trials = 200200
Number of times 4 appears = 3535
Experimental probability of rolling a 4 = 35200=0.175\frac{35}{200} = 0.175

3. A coin is flipped 50 times, and tails appear 22 times. What is the experimental probability of getting tails?
Solution:
Total number of trials = 5050
Number of times tails appear = 2222
Experimental probability of tails = 2250=0.44\frac{22}{50} = 0.44

4. A six-sided die is rolled 150 times. The number 2 appears 27 times. What is the experimental probability of rolling a 2?
Solution:
Total number of trials = 150150
Number of times 2 appears = 2727
Experimental probability = 27150=0.18\frac{27}{150} = 0.18

. A coin is tossed 80 times, and heads appear 45 times. Find the experimental probability of getting tails.
Solution:
Total number of trials = 8080
Number of times heads appear = 4545
Number of times tails appear = 8045=3580 - 45 = 35
Experimental probability of tails = 3580=0.4375\frac{35}{80} = 0.4375

6. A spinner with numbers 1 to 4 is spun 300 times, and it lands on 3 exactly 78 times. What is the experimental probability of landing on 3?
Solution:
Total trials = 300300
Number of times 3 appears = 7878
Experimental probability = 78300=0.26\frac{78}{300} = 0.26

7. A biased coin lands on heads 120 times out of 180 flips. What is the experimental probability of getting tails?
Solution:
Total flips = 180180
Heads = 120120
Tails = 180120=60180 - 120 = 60
Experimental probability of tails = 60180=0.3333\frac{60}{180} = 0.3333

8. A die is rolled 250 times, and a number 6 appears 50 times. What is the experimental probability of rolling a 6?
Solution:
Total trials = 250250
Number of times 6 appears = 5050
Experimental probability = 50250=0.2\frac{50}{250} = 0.2

9. A coin is tossed 500 times, and tails appear 230 times. Find the experimental probability of getting heads.
Solution:
Total tosses = 500500
Tails = 230230
Heads = 500230=270500 - 230 = 270
Experimental probability of heads = 270500=0.54\frac{270}{500} = 0.54

10. A die is rolled 180 times. The number 1 appears 40 times. What is the experimental probability of rolling a 1?
Solution:
Total trials = 180180
Number of times 1 appears = 4040
Experimental probability = 40180=0.2222\frac{40}{180} = 0.2222

11. A spinner has 5 equal sections numbered 1-5. It is spun 200 times, and the number 2 appears 38 times. Find the experimental probability of landing on 2.
Solution:
Total trials = 200200
Number of times 2 appears = 3838
Experimental probability = 38200=0.19\frac{38}{200} = 0.19

12. A coin is tossed 60 times, and heads appear 28 times. What is the experimental probability of getting tails?
Solution:
Total tosses = 6060
Heads = 2828
Tails = 6028=3260 - 28 = 32
Experimental probability of tails = 3260=0.5333\frac{32}{60} = 0.5333

13. A die is rolled 350 times. The number 5 appears 85 times. Find the experimental probability of rolling a 5.
Solution:
Total trials = 350350
Number of times 5 appears = 8585
Experimental probability = 85350=0.2429\frac{85}{350} = 0.2429

14. A biased die lands on 3 exactly 55 times out of 220 rolls. What is the experimental probability of rolling a 3?
Solution:
Total trials = 220220
Number of times 3 appears = 5555
Experimental probability = 55220=0.25\frac{55}{220} = 0.25

15. A coin is flipped 90 times and lands on tails 47 times. What is the experimental probability of getting tails?
Solution:
Total flips = 9090
Tails = 4747
Experimental probability of tails = 4790=0.5222\frac{47}{90} = 0.5222

16. A die is rolled 120 times, and a number 4 appears 28 times. What is the experimental probability of rolling a 4?
Solution:
Total trials = 120120
Number of times 4 appears = 2828
Experimental probability = 28120=0.2333\frac{28}{120} = 0.2333

17. A spinner is spun 400 times, and it lands on 5 exactly 92 times. Find the experimental probability of landing on 5.
Solution:
Total trials = 400400
Number of times 5 appears = 9292
Experimental probability = 92400=0.23\frac{92}{400} = 0.23

18. A coin is flipped 150 times and lands on heads 77 times. What is the experimental probability of getting heads?
Solution:
Total flips = 150150
Heads = 7777
Experimental probability = 77150=0.5133\frac{77}{150} = 0.5133

19. A die is rolled 1000 times, and the number 2 appears 180 times. Find the experimental probability of rolling a 2.
Solution:
Total trials = 10001000
Number of times 2 appears = 180180
Experimental probability = 1801000=0.18\frac{180}{1000} = 0.18

20. A six-sided die is rolled 90 times, and the number 6 appears 20 times. What is the experimental probability of rolling a 6?
Solution:
Total trials = 9090
Number of times 6 appears = 2020
Experimental probability = 2090=0.2222\frac{20}{90} = 0.2222

21. A coin is flipped 200 times, and tails appear 105 times. Find the experimental probability of getting heads.
Solution:
Total flips = 200200
Tails = 105105
Heads = 200105=95200 - 105 = 95
Experimental probability = 95200=0.475\frac{95}{200} = 0.475
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Calculation problem involving Addition and multiplication of probabilities (mutual and independent cases).

1. If P(A)=0.4P(A) = 0.4 and P(B)=0.3P(B) = 0.3, and A and B are mutually exclusive, find P(AB)P(A \cup B).
Solution:
For mutually exclusive events, P(AB)=0P(A \cap B) = 0, so:
P(AB)=P(A)+P(B)P(A \cup B) = P(A) + P(B)
=0.4+0.3=0.7= 0.4 + 0.3 = 0.7

2. If P(A)=0.5P(A) = 0.5, P(B)=0.6P(B) = 0.6, and A and B are independent, find P(AB)P(A \cap B).
Solution:
For independent events:
P(AB)=P(A)P(B)P(A \cap B) = P(A) \cdot P(B)
=0.5×0.6=0.3= 0.5 \times 0.6 = 0.3

3. If P(A)=0.25P(A) = 0.25, P(B)=0.5P(B) = 0.5, and A and B are mutually exclusive, find P(AB)P(A \cap B).
Solution:
Since A and B are mutually exclusive:
P(AB)=0P(A \cap B) = 0

4. If P(A)=0.3P(A) = 0.3, P(B)=0.4P(B) = 0.4, and they are independent, find P(AB)P(A \cup B).
Solution:
For independent events:
P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)
=0.3+0.4(0.3×0.4)= 0.3 + 0.4 - (0.3 \times 0.4)
=0.3+0.40.12=0.58= 0.3 + 0.4 - 0.12 = 0.58

5. If P(A)=0.6P(A) = 0.6, P(B)=0.2P(B) = 0.2, and they are independent, find P(AB)P(A \cap B).
Solution:
P(AB)=P(A)P(B)P(A \cap B) = P(A) \cdot P(B)
=0.6×0.2=0.12= 0.6 \times 0.2 = 0.12

6. If P(A)=0.7P(A) = 0.7 and P(B)=0.5P(B) = 0.5, and they are independent, find P(AcB)P(A^c \cap B).
Solution:
P(Ac)=1P(A)=10.7=0.3P(A^c) = 1 - P(A) = 1 - 0.7 = 0.3
Since A and B are independent:
P(AcB)=P(Ac)P(B)P(A^c \cap B) = P(A^c) \cdot P(B)
=0.3×0.5=0.15= 0.3 \times 0.5 = 0.15

7. If P(A)=0.4P(A) = 0.4 and P(B)=0.5P(B) = 0.5, and A and B are mutually exclusive, find P(AB)P(A \cup B).
Solution:
P(AB)=P(A)+P(B)P(A \cup B) = P(A) + P(B)
=0.4+0.5=0.9= 0.4 + 0.5 = 0.9

8. If P(A)=0.6P(A) = 0.6, P(B)=0.3P(B) = 0.3, and A and B are independent, find P(AcB)P(A^c \cup B).
Solution:
P(Ac)=1P(A)=10.6=0.4P(A^c) = 1 - P(A) = 1 - 0.6 = 0.4
P(AcB)=P(Ac)+P(B)P(AcB)P(A^c \cup B) = P(A^c) + P(B) - P(A^c \cap B)
=0.4+0.3(0.4×0.3)= 0.4 + 0.3 - (0.4 \times 0.3)
=0.4+0.30.12=0.58= 0.4 + 0.3 - 0.12 = 0.58

9. If P(A)=0.25P(A) = 0.25, P(B)=0.35P(B) = 0.35, and A and B are mutually exclusive, find P(AcB)P(A^c \cap B).
Solution:
P(Ac)=1P(A)=10.25=0.75P(A^c) = 1 - P(A) = 1 - 0.25 = 0.75
Since A and B are mutually exclusive,
P(AcB)=P(B)=0.35P(A^c \cap B) = P(B) = 0.35

10. If P(A)=0.55P(A) = 0.55 and P(B)=0.45P(B) = 0.45, and they are mutually exclusive, find P(AcB)P(A^c \cup B).
Solution:
P(Ac)=1P(A)=10.55=0.45P(A^c) = 1 - P(A) = 1 - 0.55 = 0.45
Since A and B are mutually exclusive,
P(AcB)=P(Ac)+P(B)P(A^c \cup B) = P(A^c) + P(B)
=0.45+0.45=0.9= 0.45 + 0.45 = 0.9

11. If P(A)=0.4P(A) = 0.4, P(B)=0.3P(B) = 0.3, and they are independent, find P(AcBc)P(A^c \cap B^c).
Solution:
P(Ac)=10.4=0.6P(A^c) = 1 - 0.4 = 0.6
P(Bc)=10.3=0.7P(B^c) = 1 - 0.3 = 0.7
Since A and B are independent,
P(AcBc)=P(Ac)P(Bc)P(A^c \cap B^c) = P(A^c) \cdot P(B^c)
=0.6×0.7=0.42= 0.6 \times 0.7 = 0.42

12. If P(A)=0.7P(A) = 0.7, P(B)=0.2P(B) = 0.2, and they are independent, find P(AcBc)P(A^c \cup B^c).
Solution:
P(Ac)=10.7=0.3P(A^c) = 1 - 0.7 = 0.3
P(Bc)=10.2=0.8P(B^c) = 1 - 0.2 = 0.8
P(AcBc)=P(Ac)+P(Bc)P(AcBc)P(A^c \cup B^c) = P(A^c) + P(B^c) - P(A^c \cap B^c)
=0.3+0.8(0.3×0.8)= 0.3 + 0.8 - (0.3 \times 0.8)
=0.3+0.80.24=0.86= 0.3 + 0.8 - 0.24 = 0.86

13. If P(A)=0.5P(A) = 0.5, P(B)=0.4P(B) = 0.4, and they are mutually exclusive, find P(ABc)P(A \cap B^c).
Solution:
P(Bc)=10.4=0.6P(B^c) = 1 - 0.4 = 0.6
Since A and B are mutually exclusive,
P(ABc)=P(A)=0.5P(A \cap B^c) = P(A) = 0.5

14. If P(A)=0.65P(A) = 0.65, P(B)=0.35P(B) = 0.35, and A and B are independent, find P(AcB)P(A^c \cap B).
Solution:
P(Ac)=10.65=0.35P(A^c) = 1 - 0.65 = 0.35
P(AcB)=P(Ac)P(B)P(A^c \cap B) = P(A^c) \cdot P(B)
=0.35×0.35=0.1225= 0.35 \times 0.35 = 0.1225

15. If P(A)=0.55P(A) = 0.55 and P(B)=0.25P(B) = 0.25, and they are mutually exclusive, find P(AcB)P(A^c \cap B).
Solution:
P(Ac)=1P(A)=10.55=0.45P(A^c) = 1 - P(A) = 1 - 0.55 = 0.45
Since A and B are mutually exclusive,
P(AcB)=P(B)=0.25P(A^c \cap B) = P(B) = 0.25

16. A bag contains 6 red balls and 4 blue balls. One ball is drawn at random. What is the probability of drawing either a red or a blue ball?
Solution:
Since all balls are either red or blue, the probability is:
P(RedBlue)=P(Red)+P(Blue)=610+410=1P(Red \cup Blue) = P(Red) + P(Blue) = \frac{6}{10} + \frac{4}{10} = 1

17. A deck of 52 playing cards is shuffled, and one card is drawn at random. What is the probability of drawing either a heart or a king?
Solution:
Hearts (13 cards) and kings (4 cards), but one of them (king of hearts) is counted twice:
P(HeartsKing)=P(Hearts)+P(King)P(HeartsKing)P(Hearts \cup King) = P(Hearts) + P(King) - P(Hearts \cap King)
=1352+452152=1652=0.3077= \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = 0.3077

18. A box contains 3 defective bulbs and 7 working bulbs. Two bulbs are drawn randomly without replacement. What is the probability that both are defective?
Solution:
First bulb is defective: 310\frac{3}{10}
Second bulb is also defective: 29\frac{2}{9}
P(BothDefective)=310×29=690=0.0667P(Both Defective) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = 0.0667

19. A school survey shows that 70% of students like math, 50% like science, and 30% like both. What is the probability that a randomly selected student likes either math or science?
Solution:
P(MathScience)=P(Math)+P(Science)P(MathScience)P(Math \cup Science) = P(Math) + P(Science) - P(Math \cap Science)
=0.7+0.50.3=0.9= 0.7 + 0.5 - 0.3 = 0.9

20. A jar has 8 green, 5 yellow, and 7 red marbles. A marble is randomly picked. What is the probability that it is green or yellow?
Solution:
P(GreenYellow)=P(Green)+P(Yellow)P(Green \cup Yellow) = P(Green) + P(Yellow)
=820+520=1320=0.65= \frac{8}{20} + \frac{5}{20} = \frac{13}{20} = 0.65

21. A die is rolled twice. What is the probability of rolling a 4 on the first roll and an even number on the second roll?
Solution:
P(4)=16,P(Even)=36=12P(4) = \frac{1}{6}, \quad P(Even) = \frac{3}{6} = \frac{1}{2}
Since the rolls are independent:
P(4Even)=P(4)×P(Even)=16×12=112P(4 \cap Even) = P(4) \times P(Even) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}

22. A class has 30 students: 18 boys and 12 girls. If a student is selected at random, what is the probability of selecting either a girl or a left-handed student, given that 5 boys and 4 girls are left-handed?
Solution:
P(Girl)=1230,P(Lefthanded)=930,P(GirlLefthanded)=430P(Girl) = \frac{12}{30}, \quad P(Left-handed) = \frac{9}{30}, \quad P(Girl \cap Left-handed) = \frac{4}{30}
P(GirlLefthanded)=P(Girl)+P(Lefthanded)P(GirlLefthanded)P(Girl \cup Left-handed) = P(Girl) + P(Left-handed) - P(Girl \cap Left-handed)
=1230+930430=1730=0.5667= \frac{12}{30} + \frac{9}{30} - \frac{4}{30} = \frac{17}{30} = 0.5667

23. Two dice are rolled. What is the probability that the sum is either 7 or 11?
Solution:
Favorable outcomes for sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
Favorable outcomes for sum = 11: (5,6), (6,5) → 2 outcomes
Total outcomes = 6×6=366 \times 6 = 36
P(711)=636+236=836=0.2222P(7 \cup 11) = \frac{6}{36} + \frac{2}{36} = \frac{8}{36} = 0.2222

24. A bag contains 5 red and 3 white balls. If two balls are picked at random, what is the probability that at least one is red?
Solution:
Complementary method:
P(BothWhite)=38×27=656=328P(Both White) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}
P(At least one Red)=1P(BothWhite)P(At\ least\ one\ Red) = 1 - P(Both White)
=1328=2528=0.8929= 1 - \frac{3}{28} = \frac{25}{28} = 0.8929

25. In a factory, 40% of products are made on Machine A and 60% on Machine B. Machine A produces 5% defective items, and Machine B produces 3% defective items. What is the probability that a randomly selected product is defective?
Solution:
Using the law of total probability:
P(Defective)=P(A)×P(DA)+P(B)×P(DB)P(Defective) = P(A) \times P(D|A) + P(B) \times P(D|B)
=(0.4×0.05)+(0.6×0.03)= (0.4 \times 0.05) + (0.6 \times 0.03)
=0.02+0.018=0.038= 0.02 + 0.018 = 0.038

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