JAMB 2018 Mathematics · Question 53

Tossing a coin and rolling a die are two separate events. What is the probability of obtaining a tail on the coin and an even number on the die?

Options

  1. A.116\frac{1}{16}
  2. B.16\frac{1}{6}
  3. C.14\frac{1}{4}
  4. D.38\frac{3}{8}

Answer: C. 14\frac{1}{4}

Explanation written with AI and checked against the answer key

Analysis

This is a problem involving independent events in probability. Two events are independent if the outcome of one does not affect the outcome of the other. Here, tossing a coin has no influence on rolling a die.

Concept

When two events, AA and BB, are independent, the probability of both occurring (the intersection) is the product of their individual probabilities:

P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
Step-by-Step Solution

Step 1: Find the probability of obtaining a tail on the coin (P(A)P(A)) A fair coin has two sides: Head (H) and Tail (T).

  • Total outcomes = 2
  • Favorable outcome (Tail) = 1
  • P(Tail)=12P(\text{Tail}) = \frac{1}{2}

Step 2: Find the probability of obtaining an even number on the die (P(B)P(B)) A standard six-sided die has the numbers {1, 2, 3, 4, 5, 6}.

  • Total outcomes = 6
  • Even numbers = {2, 4, 6}
  • Number of favorable outcomes = 3
  • P(Even)=36=12P(\text{Even}) = \frac{3}{6} = \frac{1}{2}

Step 3: Calculate the combined probability Since the events are independent, multiply the two probabilities:

P(Tail and Even)=P(Tail)×P(Even)P(\text{Tail and Even}) = P(\text{Tail}) \times P(\text{Even})
P(Tail and Even)=12×12=14P(\text{Tail and Even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
Conclusion

The probability of obtaining a tail on the coin and an even number on the die is 14\frac{1}{4}.

Exam Tip: Whenever you see the word "and" in probability questions involving independent events, it is a signal to multiply the probabilities. If you see the word "or", you generally add them.

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