A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
When a coin is tossed 3 times, the total number of equally likely outcomes in the sample space is 8. These outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
Hanif wins if all tosses give the same result, i.e., three heads (HHH) or three tails (TTT). So, the number of outcomes favourable to winning is 2. Therefore, the probability that Hanif wins is P(win) = 2/8 = 1/4. Using the complement method, the probability that Hanif will lose the game is P(lose) = 1 – P(win) = 1 – 1/4 = 3/4.
Explanation
This question tests understanding of sample space and probability calculation for equally likely outcomes. Following Example 9 from the textbook, we first determine all possible outcomes when tossing a coin 3 times. Since each toss has 2 outcomes (H or T), the total outcomes are 2³ = 8. The winning conditions (HHH or TTT) give only 2 favourable outcomes. Using the complement approach shown in Example 9 (P(E) = 1 - P(Ē)), we can efficiently calculate the losing probability as 1 - 1/4 = 3/4.
Solution Steps
Step 1: Determine the sample space. When a coin is tossed 3 times, total number of outcomes = 2³ = 8.
Step 2: List all outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
Step 3: Identify winning outcomes: HHH and TTT (2 outcomes).
Step 4: Calculate P(win) = 2/8 = 1/4.
Step 5: Calculate P(lose) = 1 - P(win) = 1 - 1/4 = 3/4.